Question

The "Jupiter Bar" is a candy bar that is only manufactured and sold in one size. Tens of thousands of bars are manufactured every day. Nutritional content appears on the bar's wrapper, including a statement that a given bar has a sodium content of 96 milligrams.
Due to variability inherent in all manufacturing, we know that some bars would have slightly less than 96 milligrams of sodium and some bars would have more than 96 milligrams of sodium even if the value of "96 milligrams" appears on the wrapper. However, there is a concern that the average sodium content in all Jupiter Bars is actually more than 96 milligrams, and we have been asked to investigate.
Below are the sodium measurements (in milligrams) from a sample of 20 Jupiter Bars:
88 93 99 104
98 103 96 99
111 90 108 98
101 112 104 102
105 101 95 104
Summary statistics:
n = 20
sample mean = 100.55
sample standard deviation = 6.304
Q = 97: median = 101: Q3 = 104
a) Test whether the sample provides evidence at the 5% level of significance that the average sodium content in all Jupiter Bars is actually more than 96 milligrams.

Answer 1

Answer:

We conclude that the average sodium content in all Jupiter Bars is actually more than 96 milligrams.

Step-by-step explanation:

We are given that there is a concern that the average sodium content in all Jupiter Bars is actually more than 96 milligrams, and we have been asked to investigate.

Summary statistics:

n = 20

sample mean = 100.55

sample standard deviation = 6.304

Q = 97: median = 101: Q3 = 104

Let $\mu$ = average sodium content in all Jupiter Bars.

So, Null Hypothesis, $H_0$ : $\mu$ = 96 milligrams     {means that the average sodium content in all Jupiter Bars is equal to 96 milligrams}

Alternate Hypothesis, $H_A$ : $\mu$ > 96 milligrams     {means that the average sodium content in all Jupiter Bars is actually more than 96 milligrams}

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 mean sodium content = 100.55

            s = sample standard deviation = 6.304

            n = sample of Jupiter bars = 20

So, the test statistics  =  $\frac{100.55-96}{\frac{6.304}{\sqrt{20} } }$  ~ $t_1_9$

                                       =  3.228

The value of t test statistics is 3.228.

Now, at 5% significance level the t table gives critical value of 1.729 at 19 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 3.228 > 1.729, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average sodium content in all Jupiter Bars is actually more than 96 milligrams.