A study showed that 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 .

Complete question is;

A study by Consumer Reports 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 52 stating that the supermarket brand was as good

as the national brand, what is the p-value?

c. At α = 0.05, what is your conclusion?

d. Should the national brand ketchup manufacturer be pleased with this conclusion?

A) H0; p =0.64. And HA; p≠0.64

B) p-value = 0.01242

C) Null hypothesis should be rejected

D) No they should not

Step-by-step explanation:

A) The hypotheses are;

Null hypothesis H0; p =0.64.

Alternative hypothesis HA; p≠0.64

B) we are given;

n = 100

x = 52

α = 0.05

The sample proportion is given by; p' = x/n = 52/100 = 0.52

Now let's check np and n(1 - p)

np = 100 x 0.64 = 64

n(1 - p) = 100(1 - 0.64) = 36

They are both greater than 5. So a two sided test

So let's find the value of the test statistic ;

z = [p' - p]/√(p(1 - p)/n)

z = [0.52 - 0.63]/√(0.64(1 - 0.64)/100)

z = - 2.5

Since 2 sided test, thus p value will be;

p = 2P(Z ≤ -2.5)

So from the table i attached,

P(Z ≤ -2.5) = 0.00621

Thus, p = 2 x 0.00621 = 0.01242

C) at α = 0.05, our p value which is 0.01242 is less than α. Thus we reject the null hypothesis.

D) There is evidence to suggest that the percentage of supermarket shoppers who believe that the supermarket Ketchup was as good as the national brand Ketchup differs from64%. Thus, the national brand Ketchup manufacturer should not be pleased with this conclusion, because the results cannot be applied to its own product.

a.

Proportion = p, Count = 53, Total = 100, Sample prop = 0.53, Std. Err = 0.046249324, z-start = -3.45951, P-value = 0.0005.

Step-by-step explanation:

Assuming 69% of supermarket shoppers believe supermarket brands to be as good as national name brands

The statistical software output for this problem is:

One sample proportion summary hypothesis test:

p : Proportion of successes

: p = 0.69

: p ≠ 0.69

Hypothesis test results:

ProportionCountTotalSample Prop.Std. Err.Z-StatP-value

p              53    100  0.530.046249324-3.459510.0005

Hence,

b) p - Value = 0.0005

c) p - value is < (0.05) so reject .

d) Since p = 0.0005, it indicates that than 0.05% of the shoppers believe the supermarket brand is as good as the name brand.