QUIZZES after midterm1 Flashcards

Question

You receive a dataset of memory performance with the following values: 6, 10, 8, 1, 16, 14. Calculate the Interquartile range (IQR).

Answer 1

INACCURATE VALUES LIE

Answer 2

CORRELATION STUDY

Answer 3

**MY ANSWERS:** a) if ppl are paid, then it's not random sampling. It's an incentive to participate. c) There is no statistical difference between the two backgrounds. d) There is a diffeence in the background scores **CORRECT ANSWERS:**

Answer 4

**MY ANSWERS:** a) pop = all college women (who can do math, maybe?) sample = just the 112 women b) indep = the confederates dep = score of the math test c) 2 groups of confederates (one is all men, other is half/half) d) ratio + continuous ( ratio has a true zero and the scores could have decimal points) e) between groups. two groups assigned to two testing groups f) not random as it was chosen from a group specifically identifying as women. should have been randomly chosen g) there's not enough data to say this fro certain h) descriptive = describes the group who identify as women (112) / inferential = it would have to be applied to the population of all the women of the college. **CORRECT ANSWERS:**

Answer 5

**MY ANSWERS:** a) pop = all college women (who can do math, maybe?) sample = just the 112 women b) indep = the confederates dep = score of the math test c) 2 groups of confederates (one is all men, other is half/half) d) ratio + continuous ( ratio has a true zero and the scores could have decimal points) e) between groups. two groups assigned to two testing groups f) not random as it was chosen from a group specifically identifying as women. should have been randomly chosen g) there's not enough data to say this fro certain h) descriptive = describes the group who identify as women (112) / inferential = it would have to be applied to the population of all the women of the college. TEACHER ANSWERS:

Answer 6

**MY ANSWERS**: a) 17 b) 2/40 = 0.05 = 5% c) 35.75 d) definitely negatively skewed -- more distributed towards the right side of the graph (more towards the higher numbers) **CORRECT ANSWERS**:

Answer 7

**MY ANSWERS:** A) it's a parameter as it describes the whole population of grade 8 students in Vancouver. B) The average score would be the 'mean', so he would need to compare against the distribution of means graph. C) the mean of the distribution would be the same as the whole population of Grade 8 students (56). D) As in, his 1 singular classroom? It would be very slightly neg scored. The distribution of means should be normally distributed as there would be many, many scores to base it off. **CORRECT ANSWERS:** a. The value of 12 is a parameter (0.5), since it has been measured for every grade 8 student in the region (the population of interest) (0.5) b. He would need to use a distribution of means, since he would like to compare the score of his 22 students (a mean) to an average score. c. The mean would be 56. The CLT tells us the mean of the distribution = the mean of the population (1 point) d. Normal. Since the sample size is 22, even if the underlying distribution is slightly skewed, due to the CLT the comparison distribution should be normal. -.25 clarify CLT makes it normal

Answer 8

D) P-hacking refers to the practice of selectively analyzing data or manipulating statistical tests in order to obtain significant results or support a desired hypothesis. By collecting multiple dependent measures but only reporting on the ones that yield significant results, the researcher is engaging in p-hacking. This selective reporting can lead to a biased presentation of the findings and can be misleading.

Answer 9

[11.12, 12.88]

Answer 10

0.33 - use Cohen's d formular: - d=M-mu/standard deviation - d=105-100/15 - =0.33

Answer 11

**C) -1.645 OR +1.645 (BECAUSE IT'S DIRECTIONAL)** For a one-tailed test with a p level of 0.05, the critical z-value can be found using the standard normal distribution. Since the test is one-tailed, we need to consider only one tail of the distribution. If the alternative hypothesis is **directional** (e.g., greater than or less than), the critical z-value will be either -1.645 or 1.645.*The alternative hypothesis is Ha and it's the 'research hypothesis'.* If the alternative hypothesis is **non-directional** (e.g., not equal to), the critical z-value will be -1.96 or 1.96. Since the question does not specify the direction of the alternative hypothesis, we assume a non-directional (two-tailed) test. Therefore, the correct answer is: -1.96 or 1.96 These critical z-values correspond to a significance level of 0.05 for a two-tailed test.Regarding the calculation of critical z-values, they are determined based on the desired significance level (α) and the type of test (one-tailed or two-tailed). **The critical z-value(s) represent the cutoff point(s) beyond which the test statistic would be considered statistically significant**. To calculate the critical z-value(s), you need to consult a standard normal distribution table

Answer 12

D) NOPE. Forgot to look at sample size of 50... CORRECT is B). Mean AND 1-tail test AND sample size allllllll increase POWER.

Answer 13

B) NOPE, NOT THE ANSWER. A) the answer? The null hypothesis typically states that there is no effect or no difference between the variables being studied. In this case, the researcher is examining whether there is a difference in the number of falls when older adults are walking while talking compared to when they are only walking. The null hypothesis assumes that there is no difference in the number of falls between the two conditions.

Answer 14

C) When we replicate findings, we become more confident in our results. (Remember, we will get a Type I error 5% of the time, or every 1 out of 20 studies we run. If a finding can't be replicated, it may well have been a Type I error).

Answer 15

**A) Since we know both the population mean and standard deviation, we can compute a z test** WRONG B) The reason for using a t-test instead of a z-test is that the population standard deviation is unknown in this case. While the *population mean* is given, the standard deviation is not specified for the entire population. Therefore, we use the sample data from the freshmen entering your college to estimate the population parameters and conduct the t-test. **The formula for the t-test is as follows:** t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))

Answer 16

C) When we are estimating a standard deviation for a population from a sample, we correct for bias (underestimation) by dividing by N-1

Answer 17

C) In this case, there is a floor effect. You can't be better than a zero response time, but you can have long response times (as might happen if your mind wanders). Thus, you would see a positive skew.

Answer 18

**C)** The t-distribution is used when the population standard deviation is unknown and is estimated using the sample data. As the sample size decreases, there is less information available to estimate the population parameters accurately. This increased uncertainty leads to wider tails in the t-distribution compared to the standard normal distribution (z-distribution), which has narrower tails. The t-distribution becomes closer to the standard normal distribution as the sample size increases, approaching the characteristics of the z-distribution.

Answer 19

D) Previously, we used the symbols SD for the standard deviation. However, that is only used when the purpose is a descriptive statistic. When we need to estimate the standard deviation based on our sample, we change the symbol to the little s. See also the section on Estimating Population Standard Deviation from a Sample

Answer 20

**a)** **TEACHER**: One way to answer this question is to look at Table B.2. What happens when you increase the degrees of freedom? **OTHER**: The other way is to understand some of the theory. If we increase our sample size, the t distribution gets narrower. That means our critical values will get closer to zero. If the degrees of freedom (df) value were increased from df = 8 to df = 20, the critical values for t would decrease. Therefore, the correct answer is a) They would decrease (move closer to zero). The t-distribution is influenced by the degrees of freedom. As the degrees of freedom increase, the t-distribution approaches the shape of the standard normal distribution (z-distribution). With a larger degrees of freedom value, the t-distribution becomes less spread out and closer to the standard normal distribution. In this case, when the degrees of freedom increase from 8 to 20, the critical values for t would decrease, moving closer to zero. This means that the t-statistic would need to be more extreme in order to reject the null hypothesis, making it more challenging to obtain statistically significant results.

Answer 21

**B)** The degrees of freedom for a single-sample t-test can be calculated by subtracting 1 from the sample size (n) because estimating the sample mean requires using one degree of freedom. In this case, with 20 participants, the degrees of freedom would be 20 - 1 = 19.

Answer 22

**A)** Confidence Interval = sample mean ± (critical value * standard error) where the critical value is obtained from the t-distribution based on the desired confidence level and degrees of freedom (sample size minus 1), and the standard error is calculated as the sample standard deviation divided by the square root of the sample size. Given: Sample mean (x̄) = 100 words per minute Sample standard deviation (s) = 50 Sample size (n) = 25 The degrees of freedom (df) for this sample is 25 - 1 = 24. First, let's calculate the standard error: Standard Error (SE) = s / √n SE = 50 / √25 SE = 10 Next, we need to determine the critical value for a 95% confidence interval with 24 degrees of freedom. Looking up the value in the t-distribution table or using a calculator, the critical value is approximately 2.064. Now we can calculate the confidence interval: Confidence Interval = x̄ ± (critical value * SE) Confidence Interval = 100 ± (2.064 * 10) **Confidence Interval = 100 ± 20.64** **Therefore, the 95% confidence interval for the sample mean reading speed is approximately [79.36 - 120.64].**

Answer 23

CORRECT **A)** WRONG **D)** **Given**: Mean of the general population (μ) = 500 Standard deviation of the general population (σ /) = 100 Sample mean (x̄) = 520 Sample standard deviation (s) = 80 Sample size (n) = 45 The standard error (SE) can be calculated as follows: **wrong** SE = s / √n **correct** SE = σ / √n Standard Error = population st.dev. / # of students of sample **wrong** SE = 80 / √45 **correct** SE = 100 / √45 STANDARD ERROR ≈ 14.91 **Therefore, Professor Milgram should use a standard error of approximately 14.91 when comparing his class's SAT scores to the general population.**

Answer 24

**D)** To calculate the standard deviation for the sample, you can use the formula for the sample standard deviation: s = √[(Σ(x - x̄)²) / (n - 1)] where: s = sample standard deviation x = individual observation x̄ = sample mean n = sample size Given the values of infant attention: 6, 6, 8, and 8, and the sample mean of 7, we can calculate the sample standard deviation. First, calculate the squared differences from the sample mean for each observation: (6 - 7)² = 1 (6 - 7)² = 1 (8 - 7)² = 1 (8 - 7)² = 1 Next, sum up these squared differences: 1 + 1 + 1 + 1 = 4 Now, divide the sum by (n - 1), where n is the sample size: 4 / (4 - 1) = 4 / 3 ≈ 1.33 Finally, take the square root of the result to obtain the sample standard deviation: s ≈ √1.33 ≈ **1.15**

Answer 25

**D)** To determine the critical cutoff for a two-tailed t-test at a p level of 0.01, we need to consider the degrees of freedom (df) and consult the t-distribution table or use a statistical calculator. In this case, a single-sample t-test is conducted on a sample of 23 people. For a single-sample t-test, the degrees of freedom are equal to the sample size minus 1: df = n - 1. Given that the sample size (n) is 23, the degrees of freedom would be 23 - 1 = 22. For a two-tailed test at a p level of 0.01, we need to split the significance level in half because we have two tails. Each tail would have an area of 0.01 / 2 = 0.005. Using the degrees of freedom (df = 22) and the tail area (0.005), we can find the critical t-value from the t-distribution table or a statistical calculator. Looking up the critical t-value for df = 22 and a tail area of 0.005, we find it to be approximately ±2.807. **Therefore, the critical cutoff for a two-tailed t-test at a p level of 0.01 would be ±2.807**.

Answer 26

**C)** To calculate the t-statistic, we can use the formula: t = (x̄ - μ) / (s / √n) where: t = t-statistic x̄ = sample mean μ = population mean s = sample standard deviation n = sample size Given: Sample mean (x̄) = 33.41 Population mean (μ) = 34.56 Sample standard deviation (s) = 4.36 Sample size (n) = 83 Substituting these values into the formula, we can calculate the t-statistic: t = (33.41 - 34.56) / (4.36 / √83) t = (-1.15) / (4.36 / √83) t = (-1.15) / (4.36 / 9.11) t = (-1.15) / 0.477 t ≈ -2.41 **Therefore, the t-statistic for this study is approximately -2.41**

Answer 27

**B)** o determine the percentile for a student who studies 8 hours a week, we need to calculate the z-score and then find the corresponding percentile from the standard normal distribution. The z-score is calculated using the formula: z = (x - μ) / σ where: z = z-score x = value we want to find the percentile for (8 hours) μ = population mean (10.82 hours) σ = population standard deviation (3.69) Substituting the given values into the formula, we can calculate the z-score: z = (8 - 10.82) / 3.69 z = -2.82 / 3.69 z ≈ -0.764 To find the percentile corresponding to this z-score, we can use a standard normal distribution table or a statistical calculator. The percentile associated with a z-score of -0.764 is approximately 22.26%. **Therefore, a student who studies 8 hours a week would be at the 22.26th percentile in the class**

Answer 28

**A)** To answer the question of whether the average salary of the 30 employees in a branch differs from that of the employees in all of the branches of the company, we would need to compare the sample mean (from the branch) to the population mean (from all branches). Given that we have information about the sample mean ($46,789), the sample standard deviation ($24,893), and the population mean ($51,256), we can use a statistical analysis to test the difference between the sample mean and the population mean. The appropriate statistical analysis to answer this question is a single-sample t-test (option A). The single-sample t-test is used when comparing the mean of a sample to a known or hypothesized population mean. In this case, we can set up the null hypothesis (H0) as: "The average salary of the 30 employees in the branch is equal to the average salary of the employees in all of the branches of the company." The alternative hypothesis (Ha) would be: "The average salary of the 30 employees in the branch is different from the average salary of the employees in all of the branches of the company." By performing a single-sample t-test, we can calculate the t-statistic using the sample mean, sample standard deviation, population mean, and the sample size (30). We can then compare the t-statistic to the critical value(s) from the t-distribution to determine if we reject or fail to reject the null hypothesis. Therefore, the correct answer is A) single-sample t test.

Answer 29

D) ---NOPE **C) CORRECT** --**In a repeated-sampling procedure, every participant gets to participate in each level of the independent variable. That means there will be no differences in things like IQ, gender, or fatigue between the groups. That reduces variability.** a) The dependent procedure generally has a larger N: Dependent sampling often involves repeated measurements or pairs of related observations, which can increase the effective sample size and provide more information for the statistical analysis. This larger sample size can lead to increased statistical power. b) Dependent procedures have higher acceptable levels of alpha: In some cases, when dependent sampling is used, the acceptable level of significance (alpha) can be adjusted to account for the dependence structure. This adjustment allows for a more lenient threshold for rejecting the null hypothesis, which can increase the power of the test. c) Dependent procedures reduce the variability in the sampling distribution: When observations are dependent, there is often a correlation or relationship between them. This dependence can reduce the variability in the sampling distribution, making it easier to detect significant differences between groups or conditions. Reduced variability can increase the power of the statistical test. Therefore, all of the options (a), (b), and (c) are correct explanations for why dependent sampling generally results in a more powerful statistical test than independent sampling.

Answer 30

**B)** In a **within-groups** design, participants encounter the dependent variable multiple times under different conditions or treatments. This repeated exposure to the dependent variable introduces the possibility of order effects, which refers to the influence that the order or sequence of the conditions may have on participants' responses. Order effects can manifest in different forms. One common type is the practice effect, where participants may improve their performance or become more familiar with the task over time, leading to biased results. Conversely, fatigue effects can occur if participants' performance deteriorates as they go through the different conditions. To mitigate order effects, researchers often implement counterbalancing techniques. Counterbalancing involves systematically varying the order of conditions across participants, such as using a Latin square or randomizing the presentation order. By doing so, researchers can control for the potential influence of order effects on the dependent variable. Therefore, the concern that arises in a within-groups design due to encountering the dependent variable twice is primarily related to order effects (option b). Compensating participants (option a), gender differences (option c), and using statistics (option d) are not direct concerns specifically associated with encountering the dependent variable twice in a within-groups design

Answer 31

**A)** In this study, participants are measured twice under two different conditions: once after consuming caffeine and once after consuming a placebo. Since the same participants are measured under both conditions, a paired-samples t test would be the most appropriate statistical test. The paired-samples t test, also known as a **dependent-samples t test or a within-subjects t test**, is used when the same participants are measured under two different conditions, treatments, or time points. It is used to determine whether there is a significant difference in the means of the paired observations. **Why not use the other tests?** A **z test** is not suitable because it is typically used when working with large sample sizes and known population parameters. An **independent-samples t test** compares the means of two independent groups, which is not applicable in this scenario. A **single-sample t test** compares the mean of a single group to a known population mean, which does not align with the study design described.

Answer 32

**C)** In a paired-samples t test, the null hypothesis assumes that there is no difference between the paired observations or conditions. Specifically, it assumes that the mean difference between the paired observations is zero. Therefore, the mean of the comparison distribution, which represents the distribution of sample mean differences under the null hypothesis, is always 0. The paired-samples t test compares the mean difference between paired observations to the mean difference expected under the null hypothesis (which is 0). If the observed mean difference significantly deviates from 0, it provides evidence to reject the null hypothesis and conclude that there is a statistically significant difference between the paired conditions

Answer 33

**A)** **In statistical hypothesis testing, a Type I error occurs when the researcher incorrectly rejects the null hypothesis when it is actually true.** In other words, it is a false positive result where the researcher concludes there is a significant effect or difference when there is none in reality

Answer 34

**B)** In a paired-samples t test, the analysis focuses on the differences between paired observations or conditions within the same participants. The goal is to determine whether there is a significant difference between the mean differences observed in the sample and the mean difference expected under the null hypothesis. To construct the comparison distribution, the mean difference is calculated for each pair of observations. The distribution of these mean differences is then used to assess the significance of the observed mean difference.

Answer 35

b) nope **D) , CORRECT** By varying the order, we can reduce order effects caused by factors such as fatigue. See the section on Next Steps in Chapter 10

Answer 36

**C)** A between-groups design, also known as an independent groups design, is a research design in which participants are assigned to different groups or conditions. Each group receives a different treatment or condition, and the groups are independent of each other. The purpose of a between-groups design is to compare the differences between the groups to determine the effect of the independent variable on the dependent variable. In the context of an independent-samples t-test, a between-groups design would involve comparing the means of two separate groups to determine if there is a significant difference between them. The t-test is used to analyze the means of two independent samples and determine whether the difference between them is statistically significant.

Answer 37

C) A between-groups design, also known as an independent groups design, is a research design in which participants are assigned to different groups or conditions. Each group receives a different treatment or condition, and the groups are independent of each other. The purpose of a between-groups design is to compare the differences between the groups to determine the effect of the independent variable on the dependent variable. In the context of an independent-samples t-test, a between-groups design would involve comparing the means of two separate groups to determine if there is a significant difference between them. The t-test is used to analyze the means of two independent samples and determine whether the difference between them is statistically significant.

Answer 38

**A)** Dr. Levine's study design involves comparing the ratings of participants before and after two different conditions (meditation and watching TV). Since the same participants are being measured under both conditions, this is a within-groups design, also known as a paired-samples design or repeated measures design.

Answer 39

**D)** In an independent-samples t test, we compare the means of two independent groups or conditions. To conduct the test, we calculate the difference between the means of the two groups and examine the distribution of these differences. This distribution represents the variability between the means of the two groups and is used to determine the statistical significance of the difference. **PROF:** Oh... wording. In this case we are using a distribution of differences between means. We are looking at one mean minus another mean. This is different from our distribution of mean differences, where we are looking at average differences between pairs.

Answer 40

**B)** Since we are working with two samples, an estimate of spread based on two samples is likely to be more accurate than an estimate of spread based on a single sample. This is why we used the pooled variance. See also Chapter 10 The Six Steps of the Independent-Samples t Test.

Answer 41

**D)** It's N-1 for both groups. So 18-1 and 18-1 equals 17 + 17 = 34

Answer 42

**D)** Since -2.365 < -0.03 < 2.365 the t-score is within the range of the critical values, FAIL TO REJECT THE NULL HYPOTHESIS

Answer 43

**B)** When the confidence interval does not include 0, it indicates that the observed difference between the means (Mx - My) is statistically significant at the specified level of significance (in this case, 0.05). This means that we can *reject the null hypothesis* and conclude that there is a significant difference between the two groups or conditions being compared.

Answer 44

**C)** Baysesian analyses use our beliefs to adjust the analysis. In this way, the analysis is much more akin to the type of decision-making we make every day. See also the Data Ethics box in Chapter 11: Beyond Hypothesis Testing.

Answer 45

**A)** In a between-groups design, different groups of participants are assigned to different conditions or treatments. In this case, the participants are divided into two groups: one group receives 70 mg of caffeine, while the other group receives a placebo drink. Each group represents a different condition or treatment.

Answer 46

**B)** To answer the question of whether the average salary of the 30 employees in the branch differs from that of the employees in all of the branches of the company, we need to compare the sample mean to the population mean. Since we are comparing one sample to a known population, the appropriate statistical analysis to use in this scenario is a single-sample t-test. A single-sample t-test is used to determine whether the mean of a sample differs significantly from a known population mean when the population standard deviation is unknown or not available. In this case, we have the sample mean ($46,789) and standard deviation ($24,893) for the 30 employees in the branch, as well as the population mean ($51,256) and standard deviation ($15,389) for all the employees in all of the tech company branches.

Answer 47

**A)** The standard error bars typically extend one standard error above and below the mean. Since the standard error is 2, we can calculate the values by adding and subtracting 2 from the mean: Upper limit: 6 + 2 = 8 Lower limit: 6 - 2 = 4 Therefore, on the bar graph, the standard error bars would begin at 4 and end at 8.

Answer 48

**A)** Mean of the comparison distribution = Experimental group mean - Control group mean Mean of the comparison distribution = 6 - 8 Mean of the comparison distribution = -2

Answer 49

**A)** Nonparametric tests are statistical analyses that do not make specific assumptions about the underlying distribution of the data. They are often used when the data do not meet the assumptions of parametric tests, such as normality or homogeneity of variances. Nonparametric tests are also useful when the sample size is small or when the data are measured on ordinal or nominal scales. In this case, the researcher is using previous knowledge of the findings to inform her statistical analysis. By leveraging existing research on the Bystander Effect, she is likely using nonparametric tests that are commonly applied in this area of study. These tests allow her to analyze the data without relying on specific assumptions about the distribution of the data.

Answer 50

**[-16.355, 0.355]** CI = (X1 - X2) ± t * SE Where: X1 and X2 are the means of group 1 and group 2, respectively. t is the critical value from the t-distribution for the desired confidence level and degrees of freedom. SE is the standard error of the difference between the means. First, let's calculate the standard error (SE) of the difference between the means: SE = sqrt((s1^2 / n1) + (s2^2 / n2)) Where: s1 and s2 are the standard deviations of group 1 and group 2, respectively. n1 and n2 are the sample sizes of group 1 and group 2, respectively. In this case, the standard error of the difference between the means (SE) is given as 5. To construct the confidence interval, we need the critical value from the t-distribution. With 30 degrees of freedom (32 + 48 - 2), a 90% confidence level corresponds to a critical value of approximately 1.697 (obtained from a t-distribution table or calculator). Now, let's calculate the confidence interval: CI = (X1 - X2) ± t * SE = (68 - 76) ± 1.697 * 5 = -8 ± 1.697 * 5 = -8 ± 8.485 = (-16.485, 0.485) Therefore, the 90% confidence interval for the difference in means between the two groups is approximately (-16.485, 0.485). Among the given options, the closest interval is [-16.355, 0.355].