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### Confidence Interval and Hypothesis Testing for GPA

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 2.87. 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 2.87 have been rejected?

#### First, find the [tex]95\%[/tex] confidence interval for the mean GPA.

The confidence interval is [tex]\square[/tex] to [tex]\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 = 2.87[/tex]
[tex]H_a: \mu \neq 2.87[/tex]

B. [tex]H_0: \mu \ \textless \ 2.87[/tex]

C. [tex]H_0: \mu = 2.87[/tex]
[tex]H_a: \mu \geq 2.87[/tex]

D. [tex]H_0: \mu \ \textgreater \ 2.87[/tex]

E. [tex]H_0: \mu = 2.87[/tex]
[tex]H_a: \mu \leq 2.87[/tex]

F. [tex]H_0: \mu \neq 2.87[/tex]
[tex]H_a: \mu = 2.87[/tex]

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

The p-value is [tex]\square[/tex].

Sagot :

GinnyAnsweridn
Let's solve this problem step by step.

### Step 1: Calculate the 95% Confidence Interval for the Mean GPA

Given Data:
- Sample size ([tex]\( n \)[/tex]) = 30
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 3.01
- Sample standard deviation ([tex]\( s \)[/tex]) = 0.24
- Critical value for a 95% confidence interval with 29 degrees of freedom (df = [tex]\( n - 1 = 29 \)[/tex]) = 2.045

Formula for Confidence Interval:
[tex]\[ \text{Confidence Interval} = \left( \bar{x} - t_\alpha \frac{s}{\sqrt{n}}, \bar{x} + t_\alpha \frac{s}{\sqrt{n}} \right) \][/tex]

Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean
- [tex]\( t_\alpha \)[/tex] is the critical value (t-score)
- [tex]\( s \)[/tex] is the sample standard deviation
- [tex]\( n \)[/tex] is the sample size

Calculation of the Margin of Error:
[tex]\[ \text{Margin of Error} = t_\alpha \times \frac{s}{\sqrt{n}} \][/tex]
[tex]\[ \text{Margin of Error} = 2.045 \times \frac{0.24}{\sqrt{30}} = 0.09 \][/tex]

Confidence Interval:
[tex]\[ \bar{x} - \text{Margin of Error} = 3.01 - 0.09 = 2.92 \][/tex]
[tex]\[ \bar{x} + \text{Margin of Error} = 3.01 + 0.09 = 3.10 \][/tex]

So, the 95% confidence interval for the mean GPA is:
[tex]\[ (2.92, 3.10) \][/tex]

### Step 2: Determine the Null and Alternative Hypotheses for a Two-Sided Test

The null and alternative hypotheses for a two-sided test at a significance level of 0.05 are:

[tex]\( H_0: \mu = 2.87 \)[/tex]
[tex]\( H_a: \mu \neq 2.87 \)[/tex]

Therefore, the correct choice is:
A. [tex]\( H_0: \mu = 2.87 \)[/tex]
[tex]\( H_a: \mu \neq 2.87 \)[/tex]

### Step 3: Calculate the Test Statistic

Formula for Test Statistic:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]

Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean
- [tex]\( \mu \)[/tex] is the hypothesized population mean
- [tex]\( s \)[/tex] is the sample standard deviation
- [tex]\( n \)[/tex] is the sample size

Calculation:
[tex]\[ t = \frac{3.01 - 2.87}{\frac{0.24}{\sqrt{30}}} = 3.18 \][/tex]

### Step 4: Calculate the p-value

The p-value for a two-sided test is calculated as:
[tex]\[ p = 2 \times (1 - \text{CDF}(|t|)) \][/tex]

Given our test statistic [tex]\( t = 3.18 \)[/tex] and degrees of freedom [tex]\( df = 29 \)[/tex]:
[tex]\[ p = 0.0035 \][/tex]

### Conclusion

- 95% Confidence Interval: [tex]\( (2.92, 3.10) \)[/tex]
- Test Statistic: [tex]\( 3.18 \)[/tex]
- p-value: [tex]\( 0.0035 \)[/tex]

Because the p-value (0.0035) is less than the significance level (0.05), we reject the null hypothesis [tex]\( H_0: \mu = 2.87 \)[/tex].

So, if a two-sided alternative had been used with a significance level of 0.05, the hypothesized mean of 2.87 would have been rejected.
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