Question

In a random sample of males, it was found that 17 write with their left hands and 212 do not. In a random sample of females, it was found that 70 write with their left hands and 439 do not. Use a 0.01 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. Identify the test statistic. 2= (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is the significance level of a = 0.01, so the null hypothesis. There 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. Click to select your answers) and then click Check Answer

Answer 1

For sample 1, we have that the sample size is N = 229, the number of favorable cases is Xi = 17, so then the sample proportion is p = x1 = 7 = 0.0742 For sample 2, we have that the sample size is N2 = 509, the number of favorable cases is X, = 70, so then the sample proportion is P2 = = 70 = 0.1375 The value of the pooled proportion is computed as p= x+8 = 229-509 = 0.1179 Also, the given significance level is a = 0.01. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: P1 = P2 Ha: P1 P2 This corresponds to a left-tailed test for which a z-test for two population proportions needs to be conducted. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a left-tailed test is ze = -2.33. The rejection region for this left-tailed testis R= {z:z < -2.33} (3) Test Statistics The Z-statistic is computed as follows: P1 - 2 P(1 - 1)(1/n1 + 1/n2) 0.0742-0.1375 == -2.467 0.1179. (1 -0.1179) (1/229 +1/509) (4) Decision about the null hypothesis Since it is observed that z = -2.467 < ze = -2.33, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0068, and since p = 0.0068 <0.01, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion p, is less than P2, at the 0.01 significance level. Confidence Interval The 99% confidence interval for pı - P2 is: -0.123 <pı - P2 < -0.004.

a)

Z =-2.47

P =0.007

The p value is less than the significance level of alpha 0.01, so reject the null hypothesis.

There is evidence to support the claim......

b) 99 Confidence interval is

(-0.123,-0.004)

Which doesn't contain zero.

Therefore null hypothesis is Rejected.

Conclusion is same as above.

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