Question

orange jelly beans is 82 and that exam scores have an approximately symmetric distribution. She gives orange jelly beans to 25 randomly selected students and finds that these students had a sample mean score of 87 with a sample standard deviation of 10. She wants to have 95% confidence in her result. 27. Conduct a hypothesis test using the confidence interval approach. Write your answer on paper and upload a photo or scan, or else put your answer in an electronic file and upload the file. In order to earn full credit your answer must include all of the following: A statement of the confidence interval in a complete sentence using the appropriate statistical symbol. Include two decimal points in your numbers. The statistical criterion you use to determine whether to reject or fail to reject the hypothesis. The result of your test in statistical terms

Answer 1

Explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 82

For the alternative hypothesis,

µ ≠ 82

This is a 2 tailed test

Since the sample mean and sample standard deviation is given, the t test would be used to determine the test statistic. The formula is

t = (x - µ)/(s/√n)

Where

x = sample mean = 87

µ = population mean = 82

s = samples standard deviation = 10

n = 25

t = (87 - 82)/(10/√25) = 2.5

α = 1 - Confidence level

α = 1 - 0.95 = 0.05

Since α = 0.05, the critical value is determined from the t distribution table.

For the left, α/2 = 0.05/2 = 0.025

For the right of 0.025 = 1 - 0.025 = 0.975

To determine the t score from the t distribution table, we would find the degree of freedom, df and look for the corresponding α value.

df = n - 1 = 25 - 1 = 24

t score = critical value = ±2.064

In order to reject the null hypothesis, the test statistic must be smaller than - 2.5 or greater than 2.5

Since - 2.064 > - 2.5 and 2.064 < 2.5, we would fail to reject the null hypothesis.

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

Margin of error = z × s/√n

Where

s = sample standard deviation

z = t score

Margin of error = 2.064 × 10/√25

= 4.13

Confidence interval = 87 ± 4.13

the lower limit of this confidence interval is

87 - 4.13 = 82.87

the upper limit of this confidence interval is

87 + 4.13 = 91.13