Answer: t= 2.032
Step-by-step explanation:
Given : Sample size : [tex]n=20[/tex]
Sample mean : [tex]\overline{x}=121[/tex]
Standard deviation : [tex]\sigma= 11[/tex]
Claim : The IQ scores of statistics professors are normally distributed, with a mean greater than 116.
Let [tex]\mu [/tex] be the mean scores of statistics professors.
Then the set of hypothesis for the given situation will be :-
[tex]H_0:\mu\leq116\\\\H_1:\mu>116[/tex]
As the alternative hypothesis is right tailed , thus the test would be right tail test.
Since the sample size is less than 30, therefore the test would be t-test .
The test statistics for the given situation will be :-
[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]\Rightarrow\ t=\dfrac{121-116}{\dfrac{11}{\sqrt{20}}}=2.03278907045\approx2.032[/tex]
Hence, the value of the test statistic : t= 2.032
Answer:
Step-by-step explanation:
To prove a hypothesis, we have to use test statisticians like the z-value which is used in normally distributed data, and this is the case.
To calculate the z-value we use: [tex]z=\frac{x-u}{\frac{o}{\sqrt{n} } }[/tex]; where x is the sample mean, u is the population mean, o is the standard deviation and n the sample size.
Replacing all values:
[tex]z=\frac{121-116}{\frac{11}{\sqrt{20} } } =\frac{5}{2.44}=2.05[/tex]
Therefore the value of the test statistic is 2.05.
(It's important to clarify that the problem isn't asking about the hypothesis, or the probability value, it's just asking for the test parameter, which in this case is just a z-value).
