Question

To test H0: ? = 4.4 versus H1: ? ? 4.4, a random sample of size n = 19 is obtained from a population that is known to be normally distributed. (a) If the sample standard deviation is determined to be s = 5.2, compute the test statistic. (b) If the researcher decides to test this hypothesis at the ? = 0.01 level of significance, use technology to determine the P-value. (c) Will the researcher reject the null hypothesis? (a) The test statistic is ?0^2 = (Round to two decimal places as needed.) (b) The P-value is (Round to three decimal places as needed.) (c) Since the P-value is than the level of significance, the researcher reject the null hypothesis H0: ? = 4.4.

          To test H0: ? = 4.4 versus H1: ? ? 4.4, a random sample of size n = 19 is obtained from a population that is known to be normally distributed.
(a) If the sample standard deviation is determined to be s = 5.2, compute the test statistic.
(b) If the researcher decides to test this hypothesis at the ? = 0.01 level of significance, use technology to determine the P-value.
(c) Will the researcher reject the null hypothesis?
(a) The test statistic is ?0^2 = 
(Round to two decimal places as needed.)
(b) The P-value is 
(Round to three decimal places as needed.)
(c) Since the P-value is 
than the level of significance, the researcher 
reject the null hypothesis H0: ? = 4.4.

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To test H0: ? = 4.4 versus H1: ? ? 4.4, a random sample of size n = 19 is obtained from a population that is known to be normally distributed.
(a) If the sample standard deviation is determined to be s = 5.2, compute the test statistic.
(b) If the researcher decides to test this hypothesis at the ? = 0.01 level of significance, use technology to determine the P-value.
(c) Will the researcher reject the null hypothesis?
(a) The test statistic is ?0^2 =
(Round to two decimal places as needed.)
(b) The P-value is
(Round to three decimal places as needed.)
(c) Since the P-value is
than the level of significance, the researcher
reject the null hypothesis H0: ? = 4.4.

Added by Sergio L.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Jon Southam

11/13/2022

Step 1

2)^2}{4.4^2} \] \[ \chi^2 = \frac{18(27.04)}{19.36} \] \[ \chi^2 = \frac{487.2}{19.36} \] \[ \chi^2 = 25.14 \] ** Show more…

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To test H0: σ = 4.4 versus H1: σ ≠ 4.4, a random sample of size n = 19 is obtained from a population that is known to be normally distributed. (a) If the sample standard deviation is determined to be s = 5.2, compute the test statistic. (b) If the researcher decides to test this hypothesis at the α = 0.01 level of significance, use technology to determine the P-value. (c) Will the researcher reject the null hypothesis? (a) The test statistic is χ0^2 = (Round to two decimal places as needed.) (b) The P-value is (Round to three decimal places as needed.) (c) Since the P-value is than the level of significance, the researcher reject the null hypothesis H0: σ = 4.4.

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Transcript

00:01 So we have a hypothesis test where the null is that sigma is equal to 4 .4.

00:07 The standard deviation is equal to 4 .4.

00:10 And then the alternative hypothesis is that it is not equal to 4 .4.

00:17 And we're told that the sample size is 19.

00:25 And we're also told that the population is normally distributed, which is nice.

00:30 And we're also told that the sample standard deviation, is 5 .2.

00:40 And we're going to do a little kye squared test to test for this hypothesis.

00:47 And that kai squared statistic is given as n minus 1 times the sample variance all over.

00:55 The value we're testing it against.

01:02 And we're also told that we're testing at the alpha of 0 .01.

01:10 Level of significance.

01:13 All right, so the first part, part a, asks us to get the test statistic.

01:19 So it's a matter of plugging our n, which is 19 minus 1 times 5 .2 squared all over 4 .4 squared, which is equal to 25 .14.

01:35 And then the p value using my spreadsheet, my google sheets.

01:48 And to do that, i use this function called chi -dist.

01:52 And in it, you put your kai squared statistic with the degrees of freedom.

01:59 And that's given as n minus 1.

02:02 So in this case, you put it in 25 .14, degrees of freedom of 18.

02:07 And the p value ends up being 0 .1211...

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