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The top food snacks consumed by adults aged 18-54 are gum, chocolate candy, fresh fruit, potato...

The top food snacks consumed by adults aged 18-54 are gum, chocolate candy, fresh fruit, potato chips, breath mints/candy, ice cream, nuts, cookies, bars, yogurt, and crackers. Out of a random sample of 26 men, 18 ranked fresh fruit in their top five snack choices. Out of a random sample of 33 women, 27 ranked fresh fruit in their top five snack choices. Is there a difference in the proportion of men and women who rank fresh fruit in their top five list of snacks?

(a-1) Choose the appropriate hypotheses. Assume ππM is the proportion of men and ππw is the proportion of women.

a. H0: πMπM – πWπW = 0vs. H1: πMπM – πWπW ≤ 0

b. H0: πMπM – πWπW ≠ 0 vs. H1: πMπM – πWπW = 0

c. H0: πMπM – πWπW = 0vs. H1: πMπM – πWπW ≥ 0

d. H0: πMπM – πWπW = 0 vs. H1: πMπM – πWπW ≠ 0

a
b
c
d

(a-2) State the decision rule for α = .02. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.)

Reject the null hypotheses if  zcalc < or zcalc > .

(b) Calculate the sample proportions. (Round your answers to 4 decimal places.)

Sample proportion   
Men         
  Women         

(c-1) Calculate the test statistic and find the p-value. (Do not round intermediate calculations. Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.)

The value of the test statistic is  and the p-value is .

(c-2) What is your conclusion?

The sample shows a significant difference in proportions.
The sample does not show a significant difference in proportions.

(d) Is normality of p1p2 assured?

No
Yes
0 0
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Answer #1

Here we have x1-18, nl=26, Pl=0.6923 X2-27,n,-33, p,-0.8182 Hypotheses are Ho:P1-P20 Level of significance α =0.04 Test is two tailed The pooled proportion is -(18+27)/(26+33)-0.762712 The standard error is: 1 ni Test Statistics: (P.-pi)-(P.-p2)--1.129 - SE Critical value: CV 2.33 Rejection Region: If z] > 2.33, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null P-value-0.259 Decision: Since p-value is not less than level of significance so we fail to reject the null Excel function for critical value Excel function for p-value: NORMSINV(1-(0.02/2)) -2* (1-NORMSDIST(1.129))

(a-1)

d. H0: πMπM – πWπW = 0 vs. H1: πMπM – πWπW ≠ 0

(a-2)

Rejection region:

If z < -2.33 or z > 2.33, reject H0

(b)

Sample proportions:

Men = 0.6923

Women = 0.8182

(c)

The test statistics is

z = -1.129

The p-value: 0.259

(c-2)

Since p-value is greater than 0.02 so we fail to reject the null hypothesis.

The sample does not show a significant difference in proportions.

(d)

Yes because number of successes and failures are both greater than 5 in both samples.

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