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Lady_Fox [76]
2 years ago
12

 A store is giving out cards labeled 1 through 10 when customers enter the store. If the card is an even​ number, you get a 15​%

discount on your purchase that day. If the card is an odd number greater than​ 6, you get a 40​% discount.​ Otherwise, you get a 25​% discount. The table shows the results of 200 customers. What is the relative frequency for each​ discount? Use pencil and paper. If the manager of the store wants approximately half of the customers to receive the 25​% ​discount, does this seem like an appropriate​ method?
Mathematics
1 answer:
liubo4ka [24]2 years ago
8 0

Answer:

yes, this made alot of sence to me and i even thought wow what teacher wrote out this problem because it was so well done!

Step-by-step explanation:

hope this helped - Willow

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A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious
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Answer:

Part1) z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191    

p_v =2*P(Z>2.191)=0.0284

Readjustment is needed

Part 2)  z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232    

z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232    

p_v =2*P(Z

No readjustment is needed

Step-by-step explanation:

Data given and notation

Part 1    

\bar X=16.32 represent the sample mean    

\sigma=0.8 represent the population standard deviation    

n=30 sample size    

\mu_o =16 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a two tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  \mu =16    

Alternative hypothesis :\mu \neq 16    

Compute the test statistic  

The statistic for this case is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191    

Give the appropriate conclusion for the test  

Since is a two tailed test the p value would be:    

p_v =2*P(Z>2.191)=0.0284

Using the value of \alpha =0.05 we see that pv and we have enough evidence to reject the null hypothesis.

Readjustment is needed

The rejection zone is given by:

(-\infty ;-1.96) , (1.96;\infty)    

Part 2

\bar X=15.82 represent the sample mean    

We can replace in formula (1) the info given like this:    

z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232    

Give the appropriate conclusion for the test  

Since is a two tailed test the p value would be:    

p_v =2*P(Z

Using the value of \alpha =0.05 we see that pv>\alpha and we have enough evidence to FAIL to reject the null hypothesis.

No readjustment is needed

The rejection zone is given by:

(-\infty ;-1.96) , (1.96;\infty)  

What action would you recommend ?

Review the procedure.

Do you reach the same conclusion ?

No we got different conclusions

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