Given : Sample size [tex]n= 189[/tex]
Sample mean : [tex]\overline{x}=7.58[/tex]
Sample standard deviation : [tex]s=1.93[/tex]
Let [tex]\mu[/tex] be the population mean of "like" ratings of male dates made by the female dates.
As per question ,
Null hypothesis : [tex]\mu\geq8.00[/tex]
Alternative hypothesis : [tex]\mu<8.00[/tex] , It means the test is a one-tailed t-test. ( we use t-test when population standard deviation is unknown.)
Test statistic:
[tex]t_{stat}=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\=\dfrac{7.58-8.00}{\dfrac{1.93}{\sqrt{189}}}=-2.99[/tex]
For 0.05 significance and df =188 (df=n-1) p-value = .001582. [By t-table]
Since .001582< 0.05
Decision: p-value < significance level , that means there is statistical significance, so we reject the null hypothesis.
Conclusion : We support the claim at 5% significance that t the population mean of such ratings is less than 8.00.
