Question

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5. The average weight is 29.0 and the standard deviation is 0.36. Assume the weight of the bags follows normal distribution. Does this provide strong evidence that the true weight is not 28.3 grams?

Answer 1

We conclude that the true mean weight is not 28.3 grams at 5% significance level.

Explanation:

We are given that some students checked 6 bags of Doritos marked with a net weight of 28.3 grams.

They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5. The average weight is 29.0 and the standard deviation is 0.36.

Let
`\mu` = true mean weight of bag.

So, Null Hypothesis,
`H_0` :
`\mu` = 28.3 grams {means that the true mean weight is 28.3 grams}

Alternate Hypothesis,
`H_A` :
`\mu`
`\\eq` 28.3 grams {means that the true mean weight is not 28.3 grams}

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. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean weight = 29.0 grams

s = sample standard deviation = 0.36 grams

n = sample of bags = 6

So, test statistics =
`(29.0-28.3)/((0.36)/(√(6) ) )` ~
`t_5`

= 4.763

The value of t test statistics is 4.763.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -2.571 and 2.571 at 5 degree of freedom for two-tailed test.

Since our test statistics does not lie within the range of critical values of t, 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 true mean weight is not 28.3 grams.