Question

A herd of 1,500 steer was fed a special high‐protein grain for a month. A random sample of 29 were weighed and had gained an average of 6.7 pounds. If the standard deviation of weight gain for the entire herd is 7.1, test the hypothesis that the average weight gain per steer for the month was more than 5 pounds.

Answer 1

By formula we know that:

z (x) = (x - m) / [sd / sqrt (n)]

where x is the value we want to know (6.7), m is the mean (5), sd is the standard deviation (7.1) and n is the sample size (29).

Replacing we have:

z (6.7) = (6.7 - 5) / [7.1 / sqrt (29)]

z = 1.289

If we look in the normal distribution table (attached), we have that the probability is 0.8997, therefore:

1 - 0.8897 = 0.1003

So the conditional probability of a herd sample earning at least 6.7 pounds per steer is 10.03%.

Now the hypothesis tells me:

m> 5

The probability is somewhat low, therefore, the most correct thing is to reject the hypothesis even though it is a fact that can occur.

Answer 2

Using the t-distribution, it is found that since the p-value is 0.1038, which is more than the standard significance level of 0.05, there is not enough evidence to reject the null hypothesis and conclude that the average weight gain per steer for the month was more than 5 pounds.

At the null hypothesis, we test if the weight gain was of at most 5 pounds, that is:

$H_0: \mu \leq 5$

At the alternative hypothesis, we test if it was more than 5 pounds, that is:

$H_1: \mu > 5$

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

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

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• s is the standard deviation of the sample.
• n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 6.7, \mu = 5, s = 7.1, n = 29$

Hence, the value of the test statistic is:

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

$t = \frac{6.7 - 5}{\frac{7.1}{\sqrt{29}}}$

$t = 1.29$

The p-value of the test is found using a right-tailed test, as we are testing if the mean is more than a value, with t = 1.29 and 29 - 1 = 28 df.

• Using a t-distribution calculator, this p-value is of 0.1038.

Since the p-value is 0.1038, which is more than the standard significance level of 0.05, there is not enough evidence to reject the null hypothesis and conclude that the average weight gain per steer for the month was more than 5 pounds.

A similar problem is given at https://brainly.com/question/13873630