Question

in a random sample of males, it was found that 23 write with their left hands and 207 do not. in a random sample of females, it was found that 61 write with their left hands and 452 do not. use a 0.05 significance level to test the claim that the rate of left - handedness among males is less than that among females. complete parts (a) through (c) below.

a. test the claim using a hypothesis test.
consider the first sample to be the sample of males and the second sample to be the sample of females. what are the null and alternative hypotheses for the hypothesis test?
a. (h_0:p_1 = p_2), (h_1:p_1>p_2)
b. (h_0:p_1
eq p_2), (h_1:p_1 = p_2)
c. (h_0:p_1 = p_2), (h_1:p_1
eq p_2)
d. (h_0:p_1geq p_2), (h_1:p_1
eq p_2)
e. (h_0:p_1leq p_2), (h_1:p_1
eq p_2)
f. (h_0:p_1 = p_2), (h_1:p_1 < p_2)
identify the test statistic.
(z=square) (round to two decimal places as needed.)
identify the p - value.
p - value=(square) (round to three decimal places as needed.)
what is the conclusion based on the hypothesis test?
the p - value is (square) the significance level of (alpha = 0.05), so (square) the null hypothesis. there (square) evidence to support the claim that the rate of left - handedness among males is less than that among females.

b. test the claim by constructing an appropriate confidence interval.
the 90% confidence interval is (square<(p_1 - p_2)<square) (round to three decimal places as needed.)
what is the conclusion based on the confidence interval?
because the confidence interval limits (square) there (square) to be a significant difference between the rates of left - handedness among males and females. there (square) evidence to support the claim that the rate of left - handedness among males is less than that among females.

c. based on the results, is the rate of left - handedness among males less than the rate of left - handedness among females?

Explanation:

Step1: Identify Hypotheses

The claim is that the rate of left - handedness among males (\(p_1\)) is less than that among females (\(p_2\)). So the null hypothesis \(H_0\) is \(p_1 = p_2\) and the alternative hypothesis \(H_1\) is \(p_1 < p_2\). Wait, but looking at the options, option F is \(H_0:p_1 = p_2\), \(H_1:p_1 < p_2\). But let's re - check. Wait, the first sample is males: \(n_1=23 + 452=475\), \(\hat{p}_1=\frac{23}{475}\). The second sample is females: \(n_2 = 61+207 = 268\), \(\hat{p}_2=\frac{61}{268}\). The claim is \(p_1 < p_2\), so \(H_0:p_1 = p_2\), \(H_1:p_1 < p_2\) which is option F? Wait no, wait the options:

Option A: \(H_0:p_1 = p_2\), \(H_1:p_1 > p_2\)

Option B: \(H_0:p_1 eq p_2\), \(H_1:p_1 = p_2\) (wrong, alternative should be different)

Option C: \(H_0:p_1 = p_2\), \(H_1:p_1 eq p_2\)

Option D: \(H_0:p_1\geq p_2\), \(H_1:p_1 eq p_2\)

Option E: \(H_0:p_1\leq p_2\), \(H_1:p_1 eq p_2\)

Option F: \(H_0:p_1 = p_2\), \(H_1:p_1 < p_2\)

Yes, the claim is \(p_1 < p_2\), so \(H_0:p_1 = p_2\), \(H_1:p_1 < p_2\) (Option F)

Step2: Calculate \(\hat{p}_1\), \(\hat{p}_2\), \(\bar{p}\)

\(n_1=23 + 452 = 475\), \(\hat{p}_1=\frac{23}{475}\approx0.0484\)

\(n_2=61 + 207=268\), \(\hat{p}_2=\frac{61}{268}\approx0.2276\)

\(\bar{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{23+61}{475 + 268}=\frac{84}{743}\approx0.1131\)

Step3: Calculate Test Statistic \(z\)

The formula for the test statistic for two - proportion test is \(z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\)

\(\hat{p}_1-\hat{p}_2=\frac{23}{475}-\frac{61}{268}\approx0.0484 - 0.2276=- 0.1792\)

\(\bar{p}(1 - \bar{p})=0.1131\times(1 - 0.1131)=0.1131\times0.8869\approx0.0999\)

\(\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{475}+\frac{1}{268}\approx0.002105+0.003731 = 0.005836\)

\(\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{0.0999\times0.005836}\approx\sqrt{0.000583}\approx0.0241\)

\(z=\frac{- 0.1792}{0.0241}\approx - 7.44\) (Wait, that seems extreme, maybe miscalculation. Wait \(n_1\): number of males who write left - handed is 23, right - handed is 452, so \(n_1 = 23+452 = 475\). \(n_2\): females left - handed 61, right - handed 207, so \(n_2=61 + 207 = 268\). \(\hat{p}_1=\frac{23}{475}\approx0.0484\), \(\hat{p}_2=\frac{61}{268}\approx0.2276\). \(\bar{p}=\frac{23 + 61}{475+268}=\frac{84}{743}\approx0.1131\). Then \(\text{SE}=\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{0.1131\times0.8869\times(\frac{1}{475}+\frac{1}{268})}\)

\(\frac{1}{475}\approx0.002105\), \(\frac{1}{268}\approx0.003731\), sum is \(0.005836\)

\(0.1131\times0.8869\approx0.0999\)

\(0.0999\times0.005836\approx0.000583\)

\(\sqrt{0.000583}\approx0.0241\)

\(z=\frac{0.0484 - 0.2276}{0.0241}=\frac{- 0.1792}{0.0241}\approx - 7.44\)

Step3: Calculate P - value

Since \(H_1:p_1 < p_2\), this is a left - tailed test. The P - value is \(P(Z < - 7.44)\). Since \(z=-7.44\) is very far in the left tail, the P - value is approximately 0.

Step4: Hypothesis Test Conclusion

Since the P - value (\(\approx0\)) is less than \(\alpha = 0.05\), we reject the null hypothesis. There is sufficient evidence to support the claim that the rate of left - handedness among males is less than that among females.

Step5: Confidence Interval for \(p_1 - p_2\) (90% confidence)

The formula for the confidence interval for \(p_1 - p_2\) is \((\hat{p}_1-\hat{p}_2)\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\)

For 90% confidence, \(z_{\alpha/2}=z_{0.05}=1.645\)

\(\hat{p}_1(1 - \hat{p}_1)=\frac{23}{475}\times(1-\frac{23}{475})=\frac{23}{475}\times\frac{452}{475}\approx0.0484\times0.9516\approx0.0460\)

\(\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}=\frac{0.0460}{475}\approx0.0000968\)

\(\hat{p}_2(1 - \hat{p}_2)=\frac{61}{268}\times(1 - \frac{61}{268})=\frac{61}{268}\times\frac{207}{268}\approx0.2276\times0.7724\approx0.1758\)

\(\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}=\frac{0.1758}{268}\approx0.000656\)

Sum of variances: \(0.0000968+0.000656 = 0.0007528\)

Standard error: \(\sqrt{0.0007528}\approx0.0274\)

\(\hat{p}_1-\hat{p}_2=\frac{23}{475}-\frac{61}{268}\approx0.0484 - 0.2276=-0.1792\)

Margin of error: \(1.645\times0.0274\approx0.0451\)

Confidence interval: \(-0.1792\pm0.0451=(-0.2243,-0.1341)\) (rounded to three decimal places)

Since the confidence interval for \(p_1 - p_2\) contains only negative values (because the upper limit is negative), this means \(p_1 - p_2<0\) or \(p_1 < p_2\), so there is sufficient evidence to support the claim.

Step6: Overall Conclusion (Part c)

Based on the results (both hypothesis test and confidence interval), the rate of left - handedness among males is less than the rate of left - handedness among females.

But let's first answer part (a) hypothesis:

Answer:

a. The correct hypotheses are F. \(H_0:p_1 = p_2\), \(H_1:p_1 < p_2\)

b. Test statistic \(z\approx - 7.44\), P - value \(\approx0\). Conclusion: Reject \(H_0\). There is sufficient evidence to support the claim.

Confidence interval: \(-0.224<(p_1 - p_2)<-0.134\) (rounded to three decimal places). Conclusion: Since the interval contains only negative values, there is sufficient evidence to support the claim.

c. Yes, the rate of left - handedness among males is less than the rate among females.

But focusing on part (a) first, the answer for part (a) is F. \(H_0:p_1 = p_2\), \(H_1:p_1 < p_2\)