Question

A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 9 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 8.5.(a) Is it appropriate to use a Student's t distribution? Explain.Yes, because the x distribution is mound-shaped and symmetric and Ï is unknown.No, the x distribution is skewed left. No, the x distribution is skewed right.No, the x distribution is not symmetric.No, Ï is known.How many degrees of freedom do we use?(b) What are the hypotheses?H0: μ = 8.5; H1: μ > 8.5H0: μ = 8.5; H1: μ â  8.5 H0: μ = 8.5; H1: μ < 8.5H0: μ < 8.5; H1: μ = 8.5H0: μ > 8.5; H1: μ = 8.5(c) Compute the t value of the sample test statistic. (Round your answer to three decimal places.)t =(d) Estimate the P-value for the test.P-value > 0.2500.100 < P-value < 0.250 0.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010(e) Do we reject or fail to reject H0?At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.(f) Interpret the results.There is sufficient evidence at the 0.05 level to reject the null hypothesis.There is insufficient evidence at the 0.05 level to reject the null hypothesis.

Answer 1

Answer:

1.) Yes, because the x distribution is mound-shaped and symmetric and σ is unknown. ;

df = 24 ;

H0 : μ = 8.5

H1 : μ ≠ 8.5 ;

1.250 ;

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

There is insufficient evidence at the 0.05 level to reject the null hypothesis.

Step-by-step explanation:

Given :

Sample size, n = 25

xbar = 9 ; Standard deviation, s = 2

α = 0.05 ;

The degree of freedom, df = n - 1 ; 25 - 1 = 24

The hypothesis (two tailed)

H0 : μ = 8.5

H1 : μ ≠ 8.5

The test statistic :

(xbar - μ) ÷ (s/√(n))

(9 - 8.5) ÷ (2/√(25))

0.5 / 0.4

Test statistic = 1.250

The Pvalue from Tscore ;

Pvalue(1.250, 24) = 0.2234

Pvalue > α ; We fail to reject H0 ;