In a random sample of males, it was found that 2828 write with their left hands
ID: 3326817 • Letter: I
Question
In a random sample of males, it was found that 2828 write with their left hands and 210210 do not. In a random sample of females, it was found that 5656 write with their left hands and 431431 do not. Use a 0.010.01 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. 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?
b. Z=
c. p=
Explanation / Answer
Solution:-
pMale = 28/238 = 0.11765
pFemale = 56/487 = 0.11499
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: PMale> PFemale
Alternative hypothesis: PMale < PFemale
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.11586
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.02531
z = (p1 - p2) / SE
z = 0.11
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.11. We use the Normal Distribution Calculator to find P(z < 0.11).
Thus, the P-value = 0.4562.
Interpret results. Since the P-value (0.4562) is greater than the significance level (0.01), we cannot reject the null hypothesis.
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