Bottles of mango juice are assumed to contain 275 milliliters of juice. There is some variation from bottle to bottle because the filling machine is not perfectly precise. Usually, the distribution of the contents is approximately Normal. An inspector measures the contents of seven randomly selected bottles from one day of production. The results are 275.4, 276.8, 273.9, 275.0, 275.8, 275.9, and 276.1 milliliters. Do these data provide convincing evidence at α = 0.05 that the mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters? (4 points)

Group of answer choices

Because the p-value of 0.0804 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.

Because the p-value of 0.0804 is greater than the significance level of 0.05, we reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters.

Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.

Because the p-value of 0.1609 is greater than the significance level of 0.05, we reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters.

Because the p-value of 1.5988 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.

Summation(x - mean)^2 = (275.4 - 275.56)^2 + (276.8 - 275.56)^2 + (273.9 - 275.56)^2 + (275.0 - 275.56)^2 + (275.8 - 275.56)^2 + (275.9 - 275.56)^2 + (276.1 - 275.56)^2 = 5.0972

Standard deviation = √(5.0972/7) = 0.85

We would set up the hypothesis test.

For the null hypothesis,

µ = 275

For the alternative hypothesis,

µ ≠ 275

This is a 2 tailed test.

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 7

Degrees of freedom, df = n - 1 = 7 - 1 = 6

t = (x - µ)/(s/√n)

Where

x = sample mean = 275.56

µ = population mean = 275

s = samples standard deviation = 0.85

t = (275.56 - 275)/(0.85/√7) = 1.74

We would determine the p value using the t test calculator. It becomes

p = 0.132

Because the p-value of 0.132 is greater than the significance level of 0.05, we would fail to reject the null hypothesis. We conclude the data does not provide convincing evidence that the mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters.

avoiddeny avoiddeny · Answer 2 · 2021-04-06T15:20:01+02:00

Answer:

C. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.

Step-by-step explanation:

I used the calculator under STAT- TESTS- 2:T-Test

μ0: 275 (mean)

x: 275.557 (add all the samples and divide by # of samples)

Sx: 0.922 (standard deviation- you can do it out or use a standard deviation calculator website- super easy, just insert the numbers you added for x above)

n: 7 (amount of samples)

μ: ≠μ (mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters)

then just press enter, and you will get a p=0.16