Math  /  Data & Statistics

Question1. A government bureau claims that more than 50%50 \% of U.S. tax returns were filed electronically last year. A random sample of 150 tax returns for last year contained 86 that were filed electronically. Test the claim at α=0.05\alpha=0.05 significance level by comparing the calculated zz-score to the critical zz-score and by comparing the p -value to the level of significance. a) State the null and alternative hypothesis. b) Calculate the critical value. c) Calculate the test statistic. d) Calculate the pp-value and compare it with the significance level. e) Can we reject the null hypothesis?

Studdy Solution

STEP 1

1. The sample size is n=150 n = 150 .
2. The number of electronically filed returns in the sample is x=86 x = 86 .
3. The significance level is α=0.05 \alpha = 0.05 .
4. The null hypothesis is that the proportion of electronically filed returns is 50% 50\% .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the critical value for the test.
3. Calculate the test statistic (z-score).
4. Calculate the p-value and compare it with the significance level.
5. Determine if we can reject the null hypothesis.

STEP 3

State the null and alternative hypotheses.
- Null Hypothesis (H0 H_0 ): p=0.50 p = 0.50 - Alternative Hypothesis (Ha H_a ): p>0.50 p > 0.50

STEP 4

Determine the critical value for a one-tailed test at α=0.05 \alpha = 0.05 .
The critical value zα z_{\alpha} for a one-tailed test at α=0.05 \alpha = 0.05 is approximately:
zα=1.645 z_{\alpha} = 1.645

STEP 5

Calculate the test statistic (z-score).
- Sample proportion (p^ \hat{p} ):
p^=xn=86150=0.5733 \hat{p} = \frac{x}{n} = \frac{86}{150} = 0.5733
- Standard error (SE SE ):
SE=p0(1p0)n=0.50×0.50150=0.0408 SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.50 \times 0.50}{150}} = 0.0408
- Test statistic (z z ):
z=p^p0SE=0.57330.500.04081.797 z = \frac{\hat{p} - p_0}{SE} = \frac{0.5733 - 0.50}{0.0408} \approx 1.797

STEP 6

Calculate the p-value and compare it with the significance level.
- The p-value for z=1.797 z = 1.797 can be found using a standard normal distribution table or calculator. It is approximately:
p-value0.036 p\text{-value} \approx 0.036

STEP 7

Determine if we can reject the null hypothesis.
- Compare the p-value with α=0.05 \alpha = 0.05 :
Since p-value=0.036<0.05 p\text{-value} = 0.036 < 0.05 , we reject the null hypothesis.
- Compare the calculated z z -score with the critical z z -score:
Since z=1.797>1.645 z = 1.797 > 1.645 , we reject the null hypothesis.
Conclusion: We reject the null hypothesis and conclude that more than 50% 50\% of U.S. tax returns were filed electronically last year.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord