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kumpel [21]
3 years ago
8

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.
Mathematics
1 answer:
Sati [7]3 years ago
7 0

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.

<u><em>Let </em></u>\mu<u><em> = population mean.</em></u>

So, Null Hypothesis, H_0 : \mu = 100.63  

Alternate Hypothesis, H_1 : \mu > 100.63  

The test statistics that would be used here <u>One-sample t test statistics</u> 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, <em><u>test statistics</u></em>  =  \frac{101.09-100.63}{\frac{0.4887}{\sqrt{7} } }  ~ t_6

                                =  2.490

The value of t test statistics is 2.490.

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