Question

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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.

Answer 1

Answer:

The correct option is;

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:

Here we have the values

μ = 275 mL

275.4

276.8

273.9

275

275.8

275.9

276.1

Sum = 1928.9

Mean (Average), = 275.5571429

Standard deviation, s = 0.921696159

We put the null hypothesis as H₀: μ₁ = μ₂

Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂

The t-test formula is as follows;

$t=\frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}}$

Plugging in the values, we have,

Test statistic = 1.599292

$t_{\alpha /2}$ at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912

Our p-value from the the test statistic = 0.1608723≈ 0.1609

Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.

Answer 2

Answer:

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:

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