Question #350842

A one sample +-test is conducted

on Ho: M = 81.6. The sample has a

mean of 84.1, s = 3.1, n = 25, and a =

01. What conclusion can be drawn?


1
Expert's answer
2022-06-16T09:49:08-0400

The following null and alternative hypotheses need to be tested:

H0:μ=81.6H_0:\mu=81.6

H1:μ81.6H_1:\mu\not=81.6

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, df=n1=24df=n-1=24 and the critical value for a two-tailed test is tc=2.79694.t_c =2.79694.

The rejection region for this two-tailed test is R={t:t>2.79694}.R = \{t:|t|>2.79694\}.

The t-statistic is computed as follows:


t=xˉμs/n=84.181.63.1/25=4.0323t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{84.1-81.6}{3.1/\sqrt{25}}=4.0323


Since it is observed that t=4.0323>2.79694=tc,|t|=4.0323>2.79694=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=24df=24 degrees of freedom, t=4.0323t=4.0323 is p=0.000486,p= 0.000486, and since p=0.000486<0.01=α,p=0.000486<0.01=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is different than 81.6, at the α=0.01\alpha = 0.01 significance level.


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