Answer:
Check the explanation
Step-by-step explanation:
This is a Dependent Sample t-test for Difference of Means. We are given subscript 1 and 2 for East and West coast fishery respectively. The hypotheses statements are
[tex]H_{0}:\mu _{1}- \mu _{2}=0~\left ( \mu _{d}=0 \right )~~~H_{A}:\mu _{1}-\mu _{2}<0~\left ( \mu _{d}<0 \right )\textup{ (Left--Tail test)}[/tex]
[tex]\textup{Given, test is at 5\% level. hence, level of significance, }\alpha =0.05[/tex]
We shall use the Student's t distribution. We assume that d has an approximately normal distribution.
The statistical analysis of data pairs gives the following values
Item AFS (East) COP (West) Difference (d d^2
Cod 19.99 17.99 2 4
Tilapi 6 13.99 -7.99 63.8401
FS 19.99 22.99 -3 9
OS 24.99 24.99 0 0
GF 29.99 19.99 10 100
Tuna 28.99 31.99 -3 9
Swordfish 23.99 23.99 0 0
Sea Bass 32.99 23.99 9 81
SB 29.99 14.99 15 225
22.01 491.8401
[tex]\sum d=22.01,~~\sum d^{2}=491.8401~\textup{ for number of data pairs, }n=9[/tex]
[tex]\textup{Average of differences, }\overline{d}=\frac{\sum d}{n}=2.445556[/tex]
[tex]\textup{Sample St. Deviation of differences, }s_{d}=\sqrt{\frac{\sum d^{2}-\frac{(\sum d)^{2}}{n}}{n-1}}[/tex]
[tex]=\sqrt{\frac{491.8401-\frac{22.01^{2}}{9}}{8}}\approx 7.3994[/tex]
[tex]\textup{Test statistic, }t=\frac{\overline{d}}{s_{d}/\sqrt{n}}=\frac{2.445556}{7.3994/\sqrt{9}}\approx 0.9915[/tex]
[tex]\textup{Degrees of freedom for }t-\textup{distribution, }df=n-1=8[/tex]
[tex]\textup{Using calculator, }p-\textup{value}=\mathbb{P}(t<0.9915)_{df=8}=0.8245[/tex]
[tex]\textup{Comparing }p-\textup{value and }\alpha \textup{ value, Fail to reject }H_{0}[/tex]
Since
The p-value is bigger than the significance level, therefore we fail to reject the null hypothesis, and will conclude with option B - There is not sufficient evidence to support the claim that west coast fish wholesalers are more expensive than east coast wholesalers.
