A study showed that 64% of supermarket shoppers believe supermarket brands to be
ID: 3365954 • Letter: A
Question
A study showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup.
A.Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 64%.
B.If a sample of 100 shoppers showed 54 stating that the supermarket brand was as good as the national brand, what is the p-value (to 4 decimals)?
C.At = .05, what is your conclusion? P-value is- greater than or equal to .05, reject greater than .05, do not reject less than or equal to .05, reject less than .05, reject equal to .05, do not reject not equal to .05
D.Should the national brand ketchup manufacturer be pleased with this conclusion? Since p = (yes or not), it indicates that than (blank) % of the shoppers believe the supermarket brand is as good as the name brand.
Explanation / Answer
Solution:-
A)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.64
Alternative hypothesis: P 0.64
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.
B) 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 ]
= 0.048
z = (p - P) /
z = - 2.083
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 -2.08 or greater than 2.08.
Thus, the P-value = 0.0376
Interpret results. Since the P-value (0.0376) is less than the significance level (0.05), we have to reject the null hypothesis.
National brand ketchup manufacturer would not be pleased with this conclusion, because the, it indicates that than 3.76 % of the shoppers believe the supermarket brand is as good as the name brand.
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