Answer:
We conclude that the mean weekly food budget for all households in this community is higher than the national average of $98.
Step-by-step explanation:
We are given that it is believed that the average amount of money spent per U.S. household per week on food is about $98, with standard deviation $10.
A random sample of 100 households in a certain affluent community yields a mean weekly food budget of $100.
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<u><em>Let </em></u><u><em> = population mean weekly food budget for all households in this community.</em></u>
SO, Null Hypothesis, : $98 {means that the mean weekly food budget for all households in this community is less than or equal to the national average of $98}
Alternate Hypothesis, : > $98 {means that the mean weekly food budget for all households in this community is higher than the national average of $98}
The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean weekly food budget = $100
= population standard deviation = $10
n = sample of households = 100
So, <u><em>test statistics</em></u> =
= 2
<em>Since in the question we are not given with the level of significance to test the hypothesis, so we assume it to be 5%. Now at 5% significance level, the z table gives </em><u><em>critical value of 1.6449</em></u><em> for right-tailed test. Since our test statistics is higher than the critical value of z as 2 > 1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that the mean weekly food budget for all households in this community is higher than the national average of $98.