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Leya [2.2K]
3 years ago
8

A quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain bra

nd of soft drink contains 120 calories as the labeling indicates. Using a random sample of 10 cans, the manager determined that the average calories per can is 124 with a standard deviation of 6 calories. At the .05 level of significance, is there sufficient evidence that the average calorie content of a 12-ounce can is greater than 120 calories? Assume that the number of calories per can is normally distributed.
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer:

We conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

Step-by-step explanation:

We are given that a quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates.

Using a random sample of 10 cans, the manager determined that the average calories per can is 124 with a standard deviation of 6 calories.

<u><em>Let </em></u>\mu<u><em> = average calorie content of a 12-ounce can.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 120 calories     {means that the average calorie content of a 12-ounce can is less than or equal to 120 calories}

Alternate Hypothesis, H_A : \mu > 120 calories     {means that the average calorie content of a 12-ounce can is greater than 120 calories}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

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

where, \bar X = sample average calories per can = 124 calories

             s = sample standard deviation = 6 calories

            n = sample of cans = 10

So, <em><u>test statistics</u></em>  =  \frac{124-120}{\frac{6}{\sqrt{10} } }  ~ t_9

                               =  2.108

The value of t test statistics is 2.108.

<em>Now, at 0.05 significance level </em><em>the t table gives critical value of 1.833 at 9 degree of freedom for right-tailed test</em><em>. Since our test statistics is more than the critical values of t as 2.108 > 1.833, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

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