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Answer:
The null and alternative hypotheses are:
[tex]H_{0}:\mu=75[/tex]
[tex]H_{a} : \mu>75[/tex]
Under the null hypothesis, the test statistic is:
[tex]t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}} }[/tex]
Where:
[tex]\bar{x} =83[/tex] is the sample mean
[tex]s=9.8206[/tex] is the sample standard deviation
[tex]n=10[/tex] is the sample size
[tex]\therefore t= \frac{83-75}{\frac{9.8206}{\sqrt{10}} }[/tex]
[tex]=2.58[/tex]
Now, we can find the right tailed t critical value at 0.01 significance level for df = n-1 = 10 - 1 = 9 using the t distribution table. The t critical value is given below:
[tex]t_{critical} =2.821[/tex]
Since the test statistic is less than the t critical value, we therefore, fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that the people do better with the new edition.
The claim that people do better with the new edition regarding the given situation is false at 0.01 level of significance.
How to form the hypotheses?
There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.
Null hypothesis is the one which researchers try to disprove.
For this case, the hypothesis we'll lay down are:
Null hypothesis: People are performing less than or equal(compared with old performance) with the new edition
Alternate hypothesis: People are doing better with the new edition
This can be symbolically written as:
Null hypothesis: [tex]H_0: \mu_2 \leq 75\\[/tex]
Alternate hypothesis: [tex]H_1: \mu_2 > 75[/tex]
(where [tex]\mu_1[/tex] = 75 is mean test score with old edition, and [tex]\mu_2[/tex] is mean test score with new edition(of the population) )
Mean score with old edition is given to be [tex]\mu_1= 75[/tex]
Mean score with new edition(of the sample) is ratio of sum of scores to total count of scores = [tex]\overline{x} = \dfrac{830}{10} = 83[/tex] ([tex]\overline{x}[/tex] is representative of the population mean with new edition)
The standard deviation for both cases is: [tex]\sigma = 10.5[/tex]
Since sample size = 10 < 30, thus, we'll perform t-test for single mean (as first mean is mean of population and not of any sample).
Degree of freedom = sample size - 1 = 9
Level of significance = 0.01
t test statistic's value for given case = [tex]\dfrac{\overline{x} - \mu_1}{s / \sqrt{n}} = \dfrac{ 83-75 }{9.82/\sqrt{10}} \approx 2.57[/tex]
(where s is sample standard deviation)
From the p-value calculator, we get the p-value for t = 2.57 at 0.01 level of significance with d.f.= 9 as : p = .015 > level of significance = 0.01
Thus, as p value is bigger than the level of significance, we fail to reject the null hypothesis, and thus, conclude that there is no significant improvement because of the new edition.
Thus, The claim that people do better with the new edition regarding the given situation is false at 0.01 level of significance.
Learn more about hypothesis testing here:
https://brainly.com/question/18831983
