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A simple random sample of 10 pages from a dictionary is obtained and the number of defined words on each page is recorded. The sample mean is found to be 53.5 definitions per page and a standard deviation of 15.7 definitions per page. To support a claim that there are more than 70,000 defined words is the same as the claim that the mean number of defined words on a page is greater than 48.0. Use a 0.05 significance level to test this claim.

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Answer:

Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean number of defined words on a page is greater than 48.0.

Step-by-step explanation:

There isn't enough statistical evidence to support the claim that the mean number of defined words on a page is greater than 48.0.

When do we use one sample t-test?

When we want to find out from the sample if the unknown population mean is different from some specific value, then we can use one sample t-test.

The test statistic is:

[tex]t = \dfrac{\overline{x} - \mu_0}{s/\sqrt{n}}[/tex]

where we have:

  • [tex]\overline{x}[/tex]= mean of the sample
  • s = sample standard deviation
  • [tex]\mu_0[/tex] = the specific value we're comparing population mean with.
  • n = sample size
  • degree of freedom = n - 1

We're given that:

  • [tex]\overline{x}[/tex]= mean of the sample = 53.5 definitions per page
  • s = sample standard deviation = 15.7 definitions per page
  • [tex]\mu_0[/tex] = the specific value we're comparing population mean with. = 48 definitions per page
  • n = sample size = 10 pages( sample of 10 pages was taken)
  • degree of freedom = n - 1 = 9
  • Level of significance = 0.05 = [tex]\alpha[/tex]
  • Want to test if population mean (the real mean number of defined words on a page) is greater than 48.0

Forming hypotheses:

  1. Null hypothesis: [tex]H_0: \mu \leq \mu_0 =48\\[/tex] (it nullifies the statement we're testing, and therefore assumes that population mean isn't greater than 48)
  2. Alternate hypothesis: [tex]H_1: \mu > \mu_0 = 48[/tex] (it favors what we want to test for)

This is a right tailed test and therefore, we accept the null hypothesis if we find [tex]t \leq t_{\alpha/2}[/tex] (where [tex]t_{\alpha/2}[/tex] is the critical value of test statistic at level of significance [tex]\alpha[/tex] and given degree of freedom).

The test statistic is:

[tex]t = \dfrac{\overline{x} - \mu_0}{s/\sqrt{n}} = \dfrac{53.5 - 48}{15.7/\sqrt{10}} \approx 1.108[/tex]

The critical value of t at 0.05 level of significance and 9 degree of freedom is [tex]t_{0.05/2} = 2.262[/tex]

Since, we got [tex]t < t_{\alpha/2}[/tex], therefore, we cannot reject the null hypothesis, and therefore, fail to accept the claim that the mean number of defined word is greater than 48.

Thus, there isn't enough statistical evidence to support the claim that the mean number of defined words on a page is greater than 48.0.

Learn more about t-test here:

https://brainly.com/question/25316952

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