A Study showed that 69% of supermarket shoppers believe supermarket brands to be
ID: 3242404 • Letter: A
Question
A Study showed that 69% 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 69%. H_0: p is H_a: p is b. If sample of 100 shoppers showed 53 stating that the supermarket brand was as good as the national brand, what is the p-value (to 4 decimals)? c. At alpha - 05, what is your conclusion? p-value is H_0 d. Should the national brand ketchup manufacturer be pleased with this conclusion? Since p =, it indicates that than % of the shoppers believe the supermarket brand is as good as the name brand.Explanation / Answer
Part-a
H0:p is =0.69
Ha; p is 0.69
Part-b
From following results p-value=0.0005
Part-c
p-value is less than 0.05,so reject H0
Part-d
No. Since pbar=0.0005, it indicates that less than 0.05% of the shoppers believe the supermarket brand is as good as the name brand
Z Test of Hypothesis for the Proportion
Data
Null Hypothesis p=
0.69
Level of Significance
0.05
Number of Successes
53
Sample Size
100
Intermediate Calculations
Sample Proportion
0.53
Standard Error
0.046249324
Z Test Statistic
-3.459510001
Two-Tail Test
Lower Critical Value
-1.959963985
Upper Critical value
1.959963985
p-Value
0.0005
Reject the null hypothesis
Z Test of Hypothesis for the Proportion
Data
Null Hypothesis p=
0.69
Level of Significance
0.05
Number of Successes
53
Sample Size
100
Intermediate Calculations
Sample Proportion
0.53
Standard Error
0.046249324
Z Test Statistic
-3.459510001
Two-Tail Test
Lower Critical Value
-1.959963985
Upper Critical value
1.959963985
p-Value
0.0005
Reject the null hypothesis
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