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Thirty GPAs from a randomly selected sample of statistics students at a college are given in the accompanying table. Assume that the population distribution is approximately normal. The technician in charge of records claimed that the population mean GPA for the whole college is 2.86. A one-sided test with a significance level of 0.05 had previously been performed on the data and the null hypothesis was rejected. Use the data to find a [tex]$95\%$[/tex] confidence interval for the mean GPA. If a two-sided alternative had been used with a significance level of 0.05, would the hypothesized mean of 2.86 have been rejected?
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First, find the [tex]$95\%$[/tex] confidence interval for the mean GPA.

The confidence interval is [tex] \square \ \square [/tex]
(Round to two decimal places as needed. Use ascending order.)

Next, test the two-sided alternative with a significance level of 0.05. Determine the null and alternative hypotheses. Choose the correct answer below.

A. [tex] H_0: \mu \ \textgreater \ 2.86 \\
H_a: \mu \leq 2.86 [/tex]

B. [tex] H_0: \# = 2.88 \\
H_a: \mu \neq 2.86 [/tex]

C. [tex] H_0: \mu / 2.86 \\
H_a: \mu = 2.85 [/tex]

D. [tex] H_0: \mu = 2.86 \\
H_a: \mu \ \textless \ 2.86 [/tex]

E. [tex] H_0: \mu = 2.86 \\
H_a: \mu \ \textgreater \ 2.86 [/tex]

F. [tex] H_0: \mu \ \textless \ 2.86 \\
H_a: \mu \geq 2.86 [/tex]

The test statistic is [tex] \square [/tex].
(Round to two decimal places as needed.)

The p-value is [tex] \square [/tex].
(Round to three decimal places as needed.)

Interpret the results of the test:

Since the p-value is [tex] \square [/tex] the significance level of 0.05, [tex] \square [/tex] [tex] H_0 [/tex]. There is [tex] \square [/tex] evidence to conclude that the mean GPA for statistics students is different from 2.86 at a significance level of 0.05.



Answer :

Sure! Let's solve this step by step:

### Step 1: Find the 95% confidence interval for the mean GPA.

Given data from the sample:
- Mean GPA ([tex]\(\bar{x}\)[/tex]): We calculated this to be 2.93 (from previously summarized data).
- Standard Deviation (s): We calculated this to be 0.18 (sample standard deviation).
- Sample Size (n): 30 (as stated).

The formula for the 95% confidence interval for the mean, when the standard deviation of the population is unknown, uses the t-distribution and is given by:

[tex]\[ \bar{x} \pm t^{*} \frac{s}{\sqrt{n}} \][/tex]

Where:
- [tex]\(t^{}\)[/tex] is the critical value from the t-distribution for 95% confidence and [tex]\(n-1\)[/tex] degrees of freedom ([tex]\(df = 29\)[/tex]).

Using the data and calculations provided:
- Critical value [tex]\(t^{
}\)[/tex]: 2.045 (typical critical value for 95% confidence with 29 df)
- Margin of error [tex]\(ME\)[/tex]: [tex]\(t^{*} \frac{s}{\sqrt{n}} = 2.045 \times \frac{0.18}{\sqrt{30}} \approx 0.07\)[/tex]

Thus, the confidence interval is:

[tex]\[ (2.93 - 0.07,\ 2.93 + 0.07) = (2.84,\ 3.01) \][/tex]

So, the 95% confidence interval is: [tex]\( \boxed{(2.84, 3.01)} \)[/tex]

### Step 2: Test the two-sided alternative with a significance level of 0.05.

Determine the null and alternative hypotheses. The correct formulation for a two-tailed test is:
- [tex]\( H_0: \mu = 2.86 \)[/tex]
- [tex]\( H_a: \mu \neq 2.86 \)[/tex]

Calculate the test statistic:

[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{2.93 - 2.86}{\frac{0.18}{\sqrt{30}}} = 1.58 \][/tex]

The test statistic value is [tex]\( \boxed{1.58} \)[/tex].

### Step 3: Determine the p-value.

For a two-tailed test, the p-value is calculated as:
- p-value = [tex]\( 2 \times P(T > |t|) \)[/tex] where T follows a t-distribution with [tex]\( df = 29 \)[/tex].

From the summarized results:
- p-value = 0.124

Thus, the p-value is [tex]\( \boxed{0.124} \)[/tex].

### Step 4: Interpret the results of the test.

Since the p-value is [tex]\(0.124\)[/tex], which is greater than the significance level ([tex]\( \alpha = 0.05 \)[/tex]), we do not reject the null hypothesis.

[tex]\( \boxed{\text{Interpretation: Since the p-value is greater than the significance level of 0.05, we do not reject } H_0. \text{ There is insufficient evidence to conclude that the mean GPA for statistics students is different from 2.86.}} \)[/tex]