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AveGali [126]
4 years ago
11

In a previous​ poll, 40​% of adults with children under the age of 18 reported that their family ate dinner together seven night

s a week. Suppose​ that, in a more recent​ poll, 456 of 1194 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ decreased? Use the alpha equals 0.01 significance level.
Mathematics
1 answer:
natali 33 [55]4 years ago
8 0

Answer:

We conclude that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ increased or remained same.

Step-by-step explanation:

We are given that in a previous​ poll, 40​% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week.

Suppose​ that, in a more recent​ poll, 456 of 1194 adults with children under the age of 18 reported that their family ate dinner together seven nights a week.

Let p = <u><em>proportion of families with children under the age of 18 who eat dinner together seven nights a week.</em></u>

SO, Null Hypothesis, H_0 : p \geq 40%      {means that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ increased or remained same}

Alternate Hypothesis, H_A : p < 40%      {means that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ decreased}

The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;

                        T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of families with children under the age of 18 who eat dinner together seven nights a week = \frac{456}{1194} = 0.38

           n = sample of adults = 1194

So, <u><em>the test statistics</em></u>  =  \frac{0.38-0.40}{\sqrt{\frac{0.38(1-0.38)}{1194} } }

                                     =  -1.424

The value of z test statistics is -1.424.

<u>Now, at 0.01 significance level the z table gives critical value of -2.326 for left-tailed test.</u>

Since our test statistic is more than the critical value of z as -1.424 > -2.326, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ increased or remained same.

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