We are given that Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month.

A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces.

Let $\mu$ = average ounces of ice cream consumed by American per month

So, Null Hypothesis, $H_0$ : $\mu \leq$ 17 ounces {means that the average American consumes less than or equal to 17 ounces of ice cream per month}

Alternate Hypothesis, $H_A$ : $\mu$ > 17 ounces {means that the average American consumes more than 17 ounces of ice cream per month}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average = 19 ounces

s = sample standard deviation = 5 ounces

n = sample of Americans = 25

So, test statistics = $\frac{19-17}{\frac{5}{\sqrt{25} } }$ ~ $t_2_4$

= 2

The value of the test statistics is 2.

Now at 0.025 significance level, the t table gives critical value of 2.06 at 24 degree of freedom for right-tailed test. Since our test statistics is less than the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the average American consumes less than or equal to 17 ounces of ice cream per month.

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4 years ago

What is the equation?

AfilCa [17]

Answer:

Step-by-step explanation: and explain

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3 years ago

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