10.2.26 -Question Help * In a survey, 34% of the respondents stated that they ta
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10.2.26 -Question Help * In a survey, 34% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected 180 pet owners and discovered that 56 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the -005 evel of significance. Click here to view the standard normal distribution table (page 1) Click here to view the standard normal distribution table (page 2) | 5% of the population size, and the sample | the Because npo (1-po) = L1 110, the sample size is requirements for testing the hypothesis satisfied (Round to one decimal place as needed.)Explanation / Answer
Solution:-
Because np0,(1-p0) = 40..39 > 10, The sample size is less than 5% of the population size, and the sample size fits the requirements for testing the hypothesis are satisfied.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.34
Alternative hypothesis: P 0.34
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.03531
z = (p - P) /
z = - 0.82
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.82 or greater than 0.82.
Thus, the P-value = 0.4122
Interpret results. Since the P-value (0.4122) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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