Question

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?

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 insufficient 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 0.0804 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 insufficient 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 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 insufficient evidence that the mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters.

Answer 1

= (275.4 + 276.8 + 273.9 + 275 + 275.8 + 275.9 + 276.1)/7 = 275.557

s = sqrt(((275.4 - 275.557)^2 + (276.8 - 275.557)^2 + (273.9 - 275.557)^2 + (275 - 275.557)^2 + (275.8 - 275.557)^2 + (275.9 - 275.557)^2 + (276.1 - 275.557)^2)/6) = 0.922 = 0.9217

H₀: = 275

H₁: 275

The test statistic t = ()/(s/)

                           = (275.557 - 275)/(0.922/)

                           = 1.60

P-value = 2 * P(T > 1.60)

           = 2 * (1 - P(T < 1.60))

           = 2 * (1 - 0.9196)

           = 0.1608

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 insufficient evidence that the mean amount of juice in all the bottles filled that day differs from the target value of 275 milliliters.