Simplify 55.55555555556 to a fraction.

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

tensa zangetsu [6.8K]

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.

Let $\mu$ = average calorie content of a 12-ounce can.

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 One-sample t test statistics 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, test statistics = $\frac{124-120}{\frac{6}{\sqrt{10} } }$ ~ $t_9$

= 2.108

The value of t test statistics is 2.108.

Now, at 0.05 significance level the t table gives critical value of 1.833 at 9 degree of freedom for right-tailed test. 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 we reject our null hypothesis.

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

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