Math / Data & Statistics

Question1. A government bureau claims that more than $50 \%$ 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 $\alpha=0.05$ significance level by comparing the calculated $z$ -score to the critical $z$ -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 $p$ -value and compare it with the significance level. e) Can we reject the null hypothesis?

Studdy Solution

STEP 1

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

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 ( $H_0$ ): $p = 0.50$ - Alternative Hypothesis ( $H_a$ ): $p > 0.50$

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

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Studdy Solution

STEP 1

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

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 ( $H_0$ ): $p = 0.50$ - Alternative Hypothesis ( $H_a$ ): $p > 0.50$

STEP 4

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

STEP 5

STEP 6

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

STEP 7

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Studdy Solution

STEP 1

1. The sample size is n=150 n = 150 n=150.2. The number of electronically filed returns in the sample is x=86 x = 86 x=86.3. The significance level is α=0.05 \alpha = 0.05 α=0.05.4. The null hypothesis is that the proportion of electronically filed returns is 50% 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 H0​): p=0.50 p = 0.50 p=0.50 - Alternative Hypothesis (Ha H_a Ha​): p>0.50 p > 0.50 p>0.50

STEP 4

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

STEP 5

STEP 6

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

STEP 7

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

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.

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

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

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