FMAL prob&stat normal and t-distributions incompl

Question 1

What is the normal distribution?

Answer 1

The normal distribution is a continuous probability distribution that is symmetric around its mean. It is commonly used in statistics to model natural variations in data. The probability density function is given by:
f(x) = (1 / (σ√(2π))) * e^(- (x - μ)² / (2σ²))
where μ is the mean and σ is the standard deviation.

Question 2

When can you use the normal distribution for confidence intervals?

Answer 2

The normal distribution can be used if:
- The population standard deviation is known.
- The sample size is large (typically n ≥ 30).
- The underlying population is normally distributed.

Question 3

What is a confidence interval?

Answer 3

A confidence interval gives a range of values within which the true population parameter is likely to fall, with a certain level of confidence.

Question 4

What is the formula for a confidence interval using the normal distribution?

Answer 4

When the population standard deviation σ is known, the confidence interval for the mean μ is:
x̄ ± Z * (σ / √n)
where x̄ is the sample mean, Z is the critical value from the standard normal distribution, and n is the sample size.

Question 5

What is the t-distribution?

Answer 5

The t-distribution (Student’s t-distribution) is a probability distribution used when estimating the mean of a normally distributed population when the sample size is small and the population standard deviation is unknown.

Question 6

When should the t-distribution be used instead of the normal distribution?

Answer 6

The t-distribution should be used when:
- The sample size is small (typically n < 30).
- The population is normally distributed.
- The population standard deviation is unknown.

Question 7

What are degrees of freedom in the t-distribution?

Answer 7

Degrees of freedom (df), denoted by ν, determine the shape of the t-distribution and are calculated as ν = n - 1, where n is the sample size.

Question 8

What is the formula for a confidence interval using the t-distribution?

Answer 8

A confidence interval for the population mean μ when the population standard deviation is unknown is:
x̄ ± t * (s / √n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t is the critical value from the t-table with (n - 1) degrees of freedom.

Question 9

Example: A sample of 15 electrical components has a mean lifespan of 489.2 hours and a standard deviation of 45.127 hours. Find a 95% confidence interval for the mean lifespan.

Answer 9

1. Sample mean: x̄ = 489.2
   
   2. Sample standard deviation: s = 45.127
   3. Sample size: n = 15, so degrees of freedom ν = 14
   4. From the t-table, for 95% confidence and 14 degrees of freedom, t = 2.145
   5. Margin of error: 2.145 * (45.127 / √15) = 25.0
   6. Confidence interval: 489.2 ± 25.0 = [464.2, 514.2]

Question 10

What is a hypothesis test using the t-distribution?

Answer 10

A hypothesis test using the t-distribution is used to determine whether the mean of a small sample significantly differs from a claimed population mean when the population standard deviation is unknown.

Question 11

What is the formula for the t-test statistic?

Answer 11

The test statistic for a one-sample t-test is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.

Question 12

Example: A manufacturer claims the average lifespan of a component is 500 hours. A sample of 15 components has a mean of 489.2 hours and a standard deviation of 45.127 hours. Test at the 5% significance level whether the lifespan is less than 500 hours.

Answer 12

1. Null hypothesis: H₀: μ = 500
   
   2. Alternative hypothesis: H₁: μ < 500 (one-tailed test)
   3. Calculate test statistic:
      t = (489.2 - 500) / (45.127 / √15) = -0.927
   4. Degrees of freedom: ν = 14, from the t-table at 5% significance, critical value = 1.761
   5. Since -0.927 > -1.761, do not reject H₀
      Conclusion: There is not enough evidence to suggest the lifespan is less than 500 hours.

Question 13

What is a paired sample t-test?

Answer 13

A paired sample t-test compares the means of two related samples, such as before-and-after measurements on the same subjects. It is used when the population standard deviation of the differences is unknown.

Question 14

What is the formula for the paired sample t-test?

Answer 14

The test statistic for a paired sample is:
t = (d̄ - μd) / (sd / √n)
where d̄ is the mean of the differences, μd is the hypothesized mean difference (often 0), sd is the standard deviation of the differences, and n is the number of pairs.

Question 15

Example: A group of 10 students took a test before and after a training program. Their score differences had a mean of 5.1 and a standard deviation of 12.44. Find a 95% confidence interval for the mean change.

Answer 15

1. Mean of differences: d̄ = 5.1
   
   2. Standard deviation of differences: sd = 12.44
   3. Sample size: n = 10, so degrees of freedom ν = 9
   4. From the t-table, for 95% confidence, t = 2.262
   5. Margin of error: 2.262 * (12.44 / √10) = 8.9
   6. Confidence interval: 5.1 ± 8.9 = [-3.8, 14.0]

Question 16

How are confidence intervals and hypothesis tests related?

Answer 16

A two-tailed hypothesis test at significance level α is equivalent to checking whether the hypothesized value of μ lies within the (1 - α) confidence interval. If the hypothesized value is outside the interval, the null hypothesis is rejected.

Question 17

Example: A manufacturer claims its product lasts 500 hours. A confidence interval for the mean lifespan is [464.2, 514.2]. Does this support the claim?

Answer 17

Since 500 lies within the confidence interval, there is no evidence to suggest the mean lifespan is different from 500 hours.