Question

Nationally, patients who go to the emergency room wait an average of 6 hours to be admitted into the hospital. Do patients at rural hospitals have a lower waiting time? The 14 randomly selected patients who went to the emergency room at rural hospitals waited an average of 4.6 hours to be admitted into the hospital. The standard deviation for these 14 patients was 2.4 hours. What can be concluded at the the a = 0.05 level of significance level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: H: VE H: v < c. The test statistic tv. (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.)

Answer 1

Solution:-

1)

a) We should use t-test for a population mean.

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

Null hypothesis: u > 6
Alternative hypothesis: u < 6

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 t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.6414
DF = n - 1

D.F = 13
c)

t = (x - u) / SE

t = - 2.183

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

The observed sample mean produced a t statistic test statistic of - 2.183.

d)

P-value = P(t < - 2.183)

Use the t-value calculator for finding p-values.

P-value = 0.0240

Interpret results. Since the P-value (0.0240) is less than the significance level (0.05), we have to reject the null hypothesis.

2)

a) We should use t-test for a population mean.

b)

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

Null hypothesis: u = 279
Alternative hypothesis: u $\neq$ 279

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 13.8152

DF = n - 1

D.F = 60
c)

t = (x - u) / SE

t = - 2.099

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than -2.099 or greater than 2.099.

d)

P-value = P(t < - 2.099) + P(t > 2.099)

Use the calculator to determine the p-values.

P-value = 0.0200

Thus, the P-value = 0.0400.

Interpret results. Since the P-value (0.0200) is greater than the significance level (0.01), we failed to reject the null hypothesis.