Mathematics, 26.04.2020 06:19, glocurlsprinces

All Fresh Seafood is a wholesale fish company based on the east coast of the U. S. Catalina Offshore Products is a wholesale fish company based on the west coast of the U. S. Table #9.2.5 contains prices from both companies for specific fish types ("Seafood online," 2013) ("Buy sushi grade," 2013). Do the data provide enough evidence to show that a west coast fish wholesaler is more expensive than an east coast wholesaler? Test at the 5% level.
Table #9.2.5: Wholesale Prices of Fish in Dollars

Fish
All Fresh Seafood Prices
Catalina Offshore Products Prices
Cod
19.99
17.99
Tilapi
6.00
13.99
Farmed Salmon
19.99
22.99
Organic Salmon
24.99
24.99
Grouper Fillet
29.99
19.99
Tuna
28.99
31.99
Swordfish
23.99
23.99
Sea Bass
32.99
23.99
Striped Bass
29.99
14.99

Answers: 1

Show answers

Answers

answered: jaylaa04

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

$H_{0}:\mu _{1}- \mu _{2}=0~\left ( \mu _{d}=0 \right )~~~H_{A}:\mu _{1}-\mu _{2}$

$\textup{Given, test is at 5\% level. hence, level of significance, }\alpha =0.05$

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 00

GF 29.99 19.99 10 100

Tuna 28.99 31.99 -3 9

Swordfish23.99 23.99 00

Sea Bass32.99 23.99 981

SB 29.99 14.99 15 225

22.01 491.8401

$\sum d=22.01,~~\sum d^{2}=491.8401~\textup{ for number of data pairs, }n=9$

$\textup{Average of differences, }\overline{d}=\frac{\sum d}{n}=2.445556$

$\textup{Sample St. Deviation of differences, }s_{d}=\sqrt{\frac{\sum d^{2}-\frac{(\sum d)^{2}}{n}}{n-1}}$

$=\sqrt{\frac{491.8401-\frac{22.01^{2}}{9}}{8}}\approx 7.3994$

$\textup{Test statistic, }t=\frac{\overline{d}}{s_{d}/\sqrt{n}}=\frac{2.445556}{7.3994/\sqrt{9}}\approx 0.9915$

$\textup{Degrees of freedom for }t-\textup{distribution, }df=n-1=8$

$\textup{Using calculator, }p-\textup{value}=\mathbb{P}(t$

$\textup{Comparing }p-\textup{value and }\alpha \textup{ value, Fail to reject }H_{0}$

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.

answered: Guest

step-by-step explanation:

answered: Guest

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.