A dietician read in a survey t
A dietician read in a survey that 63.3% of adults in the U.S. donot eat breakfast at least 3 days a week. She believes that theproportion that skip breakfast 3 days a week is different than0.633. To verify her claim, she selects a random sample of 73adults and asks them how many days a week they skip breakfast. 48of them report that they skip breakfast at least 3 days a week.Test her claim at αα = 0.05.The correct hypotheses would be:
- H0:p≤0.633H0:p≤0.633HA:p>0.633HA:p>0.633 (claim)
- H0:p≥0.633H0:p≥0.633HA:p<0.633HA:p<0.633 (claim)
- H0:p=0.633H0:p=0.633HA:p≠0.633HA:p≠0.633 (claim)
Since the level of significance is 0.05 the critical value is 1.96and -1.96The test statistic is: (round to 3 places)The p-value is: (round to 3 places)The decision can be made to:
- reject H0H0
- do not reject H0H0
The final conclusion is that:
- There is enough evidence to reject the claim that theproportion that skip breakfast 3 days a week is different than0.633.
- There is not enough evidence to reject the claim that theproportion that skip breakfast 3 days a week is different than0.633.
- There is enough evidence to support the claim that theproportion that skip breakfast 3 days a week is different than0.633.
- There is not enough evidence to support the claim that theproportion that skip breakfast 3 days a week is different than0.633.
Answer:
Solution :
Given that,
= 0.633
1 – = 0.367
n = 73
x = 48
Level of significance = 
= 0.05
Point estimate = sample proportion = 
= x / n = 0.658
This a two- tailed test.
The null and alternative hypothesis is,
Ho: p = 0.633
Ha: p 
0.633 ( claim )
Critical value of the significance level isα = 0.05, and the critical value for a two-tailed testis
= 1.96
Test statistics
z = (– ) / 
*(1-)/ n
= ( 0.658 – 0.633) / (0.633*0.367) / 73
= 0.443
P-value = 2 * P(Z > z )
= 2 * ( 1 – P(Z < 0.443 ))
= 2 * 0.3289
= 0.658
The p-value is p = 0.658, and since p = 0.658 > 0.05, it isconcluded that fail to reject the null hypothesis.
Do not reject H0.
There is not enough evidence to reject the claim thatthe proportion that skip breakfast 3 days a week is different than0.633.
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