Math  /  Data & Statistics

Question- Hypothesis Tests for a Question 8, 10.3.33-P HW Score: 97.62%,8.7997.62 \%, 8.79 of 9 points Part 6 of 9 Points: 0.79 of 1 Save
A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the College Board, scores are normally distributed with μ=518\mu=518. The teacher obtains a random sample of 2200 students, puts them through the review class, and finds that the mean math score of the 2200 students is 523 with a standard deviation of 118 . Complete parts (a) through (d) below.
Click the icon to view the t-Distribution Area in the Right Tail. A. No, because every increase in score is practically significant. B. Yes, because the score became more than 0.97%0.97 \% greater. C. No, because the score became only 0.97%0.97 \% greater D. Yes, because every increase in score is practically significant. (d) Test the hypothesis at the α=0.10\alpha=0.10 level of significance with n=400n=400 students. Assume that the sample mean is still 523 and the sample standard deviation is still 118 . Is a sample mean of 523 significantly more than 518 ? Conduct a hypothesis test using the P-value approach.
Find the test statistic. t0=\mathrm{t}_{0}= \square (Round to two decimal places as needed.) vv an example Get more help - Clear all Check answer 3 Nov 24 10:09 INTL 붕

Studdy Solution

STEP 1

1. The null hypothesis H0 H_0 is that the mean score μ=518\mu = 518.
2. The alternative hypothesis Ha H_a is that the mean score μ>518\mu > 518.
3. The sample size n=400 n = 400 .
4. The sample mean xˉ=523\bar{x} = 523.
5. The sample standard deviation s=118 s = 118 .
6. The significance level α=0.10\alpha = 0.10.

STEP 2

1. Calculate the test statistic.
2. Determine the critical value or P-value.
3. Make a decision to reject or fail to reject the null hypothesis.

STEP 3

Calculate the test statistic using the formula for the t-statistic:
t0=xˉμs/nt_0 = \frac{\bar{x} - \mu}{s / \sqrt{n}}
Substitute the given values:
t0=523518118/400t_0 = \frac{523 - 518}{118 / \sqrt{400}}

STEP 4

Calculate the denominator:
s/n=118/400=118/20=5.9s / \sqrt{n} = 118 / \sqrt{400} = 118 / 20 = 5.9

STEP 5

Calculate the test statistic:
t0=5235185.9=55.90.85t_0 = \frac{523 - 518}{5.9} = \frac{5}{5.9} \approx 0.85

STEP 6

Determine the P-value using the t-distribution table or calculator for t0=0.85 t_0 = 0.85 with n1=399 n - 1 = 399 degrees of freedom.

STEP 7

Since n n is large, use the standard normal distribution as an approximation. Find the P-value for t0=0.85 t_0 = 0.85 .

STEP 8

Compare the P-value to the significance level α=0.10\alpha = 0.10. If the P-value is less than α\alpha, reject the null hypothesis.

STEP 9

The P-value is greater than α=0.10\alpha = 0.10, so we fail to reject the null hypothesis.
The test statistic is:
t0=0.85\mathrm{t}_{0} = 0.85

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