Answer:
a
[tex]t = 1.9166[/tex]
b
The nutritionist does not have sufficient evidence to reject the writers claim
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 75[/tex]
The sample size is [tex]n = 20[/tex]
The sample mean is [tex]\= x = 78[/tex]
The standard deviation [tex]\sigma = 7[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
Now the
Null Hypothesis is [tex]H_o : \mu = 75[/tex]
Alternative Hypothesis is [tex]\mu \ne 75[/tex]
generally the degree of freedom is mathematically represented as
[tex]df = n-1[/tex]
[tex]df = 20 -1[/tex]
[tex]df = 19[/tex]
For a significance level of 0.05 and 19 degrees of freedom, the critical value for the t-test is 2.093. This is obtain from the t-distribution table
The test statistic is mathematically evaluated as
[tex]t = \frac{ \= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{78 - 75 }{\frac{7 }{\sqrt{20} } }[/tex]
[tex]t = 1.9166[/tex]
Since the test statistic is below the critical value then the Null hypothesis can not be rejected
So we can conclude that the nutritionist does not have sufficient evidence to reject the writers claim
