In testing H₀: µ = 23 versus H_A: µ 23 when =26,s = 6,and n = 20,what is the value of the t-statistic? A)10.00 B)2.24 C)-2.24 D)1.34

Question

In testing H₀: µ = 23 versus H_A: µ > 23 when =26,s = 6,and n = 20,what is the value of the t-statistic? A)10.00 B)2.24 C)-2.24 D)1.34

Quizplus · Accepted Answer

First, we must calculate the test statistic using the formula:

t = (x̄ - μ) / (s / √n)

where x̄ is the sample mean, μ is the hypothesized mean, s is the sample standard deviation, and n is the sample size.

Plugging in the values given, we get:

t = (26 - 23) / (6 / √20) = 2.24

Since the alternative hypothesis is one-tailed (i.e. μ > 23), we are interested in the positive tail of the t-distribution. Looking up the t-distribution table with degrees of freedom (df) = n - 1 = 19 and significance level α = 0.05, we find that the critical value is 1.734. Since our calculated t-value (2.24) is greater than the critical value (1.734), we would reject the null hypothesis and conclude that there is evidence to support the claim that the true mean is greater than 23.

In Testing H0: µ = 23 Versus HA: µ >

In Testing H₀: µ = 23 Versus H_A: µ >