Answer to Question #202170 in Statistics and Probability for Delmundo

Question

Read and analyze the given problem. Provide a complete solution, Table, and a graph corresponds to the answer.

According to last year’s report, a filipino household spends an average of P400/day for food. Suppose you took a sample of 25 households. You determined how much each household spent for the food daily and the results revealed a mean of P380 and a standard deviation of P21.50. with 99% confidence what would be your conclusion

Due to ASF issue it was aired that the average price of a kilo of pork in MM is P195.00. However, a sample of 15 prices randomly collected from different markets in MM showed an average of P200.00 and with standard deviation of P9.50. using 0.05 significant level. What would be your conclusion for the price of pork in MM.

Answer 1

answer 1

given:

$\mu =P400, \space n=25,\space \bar x =P380,\space \sigma =P21.50\\$

since we are given population standard deviation, we use z-test

step1:

state the hypothesis and identify the claim

$H_0 : \mu \leq P400\\ H_1: \mu > P400, claim(one-tailed \space right)$

stpe 2:

confidence is $=99\%$

step 3:

determine the critical value using the table

$z_{critical}=+2.576$

step 4:

compute the one sample z test value using the formula

$z_{computed}=\frac{\bar x- \mu}{\frac{\sigma }{\sqrt{n}}}\\ z_{computed}=\frac{380- 400}{\frac{21.50}{\sqrt{25}}}\\ z_{computed}=\frac{-20}{\frac{21.50}{\sqrt{25}}}\\ z_{computed}=\frac{-20}{21.50} {\sqrt{25}}\\ z_{computed}=-4.6511$

step 5:

decision rule ,compare the computed and critical value of z

$-4.6511<2.576 \\ accept\space H_0 \space and \space Reject \space H_1$

step 6;

so conclusion is household spent for the food daily less than P400

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answer 2

given:

$\mu =P195, \space n=15,\space \bar x =P200,\space \sigma =P9.50\\$

since we are given population standard deviation, we use z-test

step1:

state the hypothesis and identify the claim

$H_0 : \mu \leq P195\\ H_1: \mu > P195, claim(one-tailed \space right)$

stpe 2:

the level of significance is $\alpha =0.05$

step 3:

determine the critical value using the table

$z_{critical}=+1.645$

step 4:

compute the one sample z test value using the formula

$z_{computed}=\frac{\bar x- \mu}{\frac{\sigma }{\sqrt{n}}}\\ z_{computed}=\frac{200- 195}{\frac{9.50}{\sqrt{15}}}\\ z_{computed}=\frac{5}{\frac{9.50}{\sqrt{15}}}\\ z_{computed}=\frac{5}{9.50} {\sqrt{15}}\\ z_{computed}=2.038$

step 5:

decision rule ,compare the computed and critical value of z

$2.038>1.645 \\ Reject \space H_ 0\space ​ and\space accept H _1 ​$

step 6;

so conclusion is  price of pork in MM more than P195