Question

Use a one-sample t ‑test, based on the data below, to test the null hypothesis H0:µ=100.63 against the alternative hypothesis H1:µ>100.63 . The sample has a mean of x⎯⎯⎯=101.09 and a standard deviation of s=0.4887 . 100.68,101.23,100.82,101.15,100.96,100.70,102.09 Calculate the standard error (SE) and the t ‑statistic for this test. Give the standard error to four decimal places and t to three decimal places.

Answer 1

Answer:

The standard error (SE) is 0.1847.

The t-statistic for this test is 2.490.

Step-by-step explanation:

We are given that the sample has a mean of $\bar X$ = 101.09 and a standard deviation of s = 0.4887 .

Also, the 7 sample values are also given.

Let $\mu$ = population mean.

So, Null Hypothesis, $H_0$ : $\mu$ = 100.63  

Alternate Hypothesis, $H_1$ : $\mu$ > 100.63  

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

                         T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean = 101.09

            s = sample standard deviation = 0.4887

            n = sample values = 7

The Standard Error (SE) is given by =  $\frac{s}{\sqrt{n} }$  =  $\frac{0.4887}{\sqrt{7} }$ = 0.1847

So, test statistics  =  $\frac{101.09-100.63}{\frac{0.4887}{\sqrt{7} } }$  ~ $t_6$

                                =  2.490

The value of t test statistics is 2.490.