Question

16.
A random and independently chosen sample of four bags of horse carrots, each bag labeled 20 pounds, had weights of 20.5, 19.9, 20.9, and 20.0 pounds. Assume that the distribution of weights in the population is Normal Complete parts a through c below. a. Test the hypothesis that the population mean weight is not 20 pounds Which of the following correctly states Ho and Ha? H0 : μ > 20 Ha : μ 20 Ho: μ = 20 Ha : μ > 20 Ho : μ 20 Ha : μ = 20 Ho : μ = 20 Ha : μ 20 H0 : μ-20 Ha : μ < 20 E. Ha : μ < 20 Find the test statistic. (Type an integer or decimal rounded to two decimal places as needed.) Find the p-value (Type an integer or decimal rounded to three decimal places as needed.)
A random and independently chosen sample of four bags of horse​ carrots, each bag labeled 20​ pounds, had weights of

20.520.5​,

19.919.9​,

20.920.9​,

and

20.020.0

pounds. Assume that the distribution of weights in the population is Normal. Complete parts a through c below.

Answer 1

Solution:-

Data Mean Standard Error Median Mode Standard Deviation0.464579 Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 20.325 0.232289 20.25 0.215833 2.47814 0.560977 19.9 20.9 81.3 4

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ = 20
Alternative hypothesis: μ ≠ 20

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.232289

z = (x - μ) / SE

z = 1.399

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic having less than -1.399 or greater than 1.399.

Thus, the P-value = 0.162.

Interpret results. Since the P-value (0.162) is greater than the significance level (0.05), we cannot reject the null hypothesis.

(B) Do not reject H0, there is no reason to believe that the population mean is not 20 pounds on the basis of these data at a significance level of 0.05.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ > 20
Alternative hypothesis: μ < 20

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.232289

z = (x - μ) / SE

z = 1.399

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 1.399

Thus the P-value in this analysis is 0.92

Interpret results. Since the P-value (0.92) is greater than the significance level (0.05), we cannot reject the null hypothesis.

A) Do not reject H0, there is no reason to believe that the population mean is less than 20 pounds on the basis of these data at a significance level of 0.05.

c)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ < 20
Alternative hypothesis: μ > 20

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.232289

z = (x - μ) / SE

z = 1.399

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 1.399

Thus the P-value in this analysis is 0.081.

Interpret results. Since the P-value (0.081) is greater than the significance level (0.05), we cannot reject the null hypothesis.

A) Do not reject H0, there is no reason to believe that the population mean is more than 20 pounds on the basis of these data at a significance level of 0.05.