Question

A dietician read in a survey that 63.3% of adults in the U.S. do not eat breakfast at least 3 days a week. She believes that the proportion that skip breakfast 3 days a week is different than 0.633. To verify her claim, she selects a random sample of 73 adults and asks them how many days a week they skip breakfast. 48 of 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≤0.633
  HA:p>0.633HA:p>0.633 (claim)
• H0:p≥0.633H0:p≥0.633
  HA:p<0.633HA:p<0.633 (claim)
• H0:p=0.633H0:p=0.633
  HA:p≠0.633HA:p≠0.633 (claim)

Since the level of significance is 0.05 the critical value is 1.96 and -1.96

The 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 the proportion that skip breakfast 3 days a week is different than 0.633.
• There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.633.
• There is enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.633.
• There is not enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.633.

Answer 1

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 is α = 0.05, and the critical value for a two-tailed test is

= 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 is concluded that fail to reject the null hypothesis.

Do not reject H0.

There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.633.