A brand of powdered milk is advertised as having a net weight of 250 grams. A curious consumer obtained the net weight of 10 randomly selected cans. The values obtained are: 245, 248, 242, 245, 246, 248, 250, 245, 243 and 249 grams. Is there reason to believe that the average net weight of the powdered milk cans is less than 250 grams at 5% level of significance? Assume the net weight is normally distributed with unknown population variance.
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Answer:
15. B.
16. C.
17. A.
18. B.
19. D.
20. A.
Explanation:
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Step-by-step explanation:
To determine if there is reason to believe that the average net weight of the powdered milk cans is less than 250 grams at 5% level of significance, we can perform a one-sample t-test.
The null hypothesis is that the true mean net weight of the powdered milk cans is equal to 250 grams, while the alternative hypothesis is that it is less than 250 grams.
To perform the test, we first calculate the sample mean and sample standard deviation of the net weights:
Sample mean:
$\bar{x} = \frac{245 + 248 + 242 + 245 + 246 + 248 + 250 + 245 + 243 + 249}{10} = 246.1$
Sample standard deviation:
$s = \sqrt{\frac{\sum_{i=1}^{10} (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{(245-246.1)^2 + (248-246.1)^2 + \cdots + (249-246.1)^2}{9}} \approx 2.32$
The test statistic is given by:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
where $\mu_0$ is the hypothesized population mean (250 grams), $s$ is the sample standard deviation, and $n$ is the sample size (10).
Substituting the values, we get:
$t = \frac{246.1 - 250}{2.32/\sqrt{10}} \approx -2.57$
Using a t-distribution table with 9 degrees of freedom (10-1), the critical value for a one-tailed test with 5% level of significance is -1.833. Since the calculated test statistic is less than the critical value, we reject the null hypothesis and conclude that there is evidence to suggest that the average net weight of the powdered milk cans is less than 250 grams at 5% level of significance.