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nata0808 [166]
3 years ago
12

The information systems department of your firm has developed a new database program to store critical information on point of s

ale purchases. The purchases are made during a busy time of day when the computer is performing many tasks so the amount of computer time used for each transaction is very critical. The old database program would take an average of 55 milliseconds to retrieve one piece of data. The IS department hopes the new program will reduce this time to retrieve data. The new program was tested during a typical day. The number of data retrievals was 144 and the average time to retrieve a piece of data was 53 milliseconds with a sample standard deviation of 16 milliseconds.
a) State the null and alternative hypothesis.
b) Calculate the test statistic.
c) Using the P-value approach in your decision making, at the 0.05 level of significance, is there evidence that the new computer program has reduced the time to retrieve the data? Report p-value and use it to support your conclusion.
d) Does your conclusion in part c) change if an a of 0.025 is used?
Mathematics
1 answer:
artcher [175]3 years ago
4 0

Step-by-step explanation:

x'=53

σ=16

n=144

a) hypothesis

H0:µ=55

Ha:µ<55

This is a left tailed test

b) Test statistics

  • z=(x'-u)/(sigma/sqrt {n})

=(53-55)/(16/sqrt{144})

=-1.5

c)Pvalue at z=|-1.5|

pvalue= p(z>1.5)

=1-0.933193

=0.066807

=0.0668

<u>Decision</u>

since pvalue>alpha(0.05) fail to reject the null hypotheis.

<u>Conclusion</u>

There is not sufficient evidence to support the claim the new computer program has not reduce the time to retrieve the data.

d)since Pvalue>alpha(0.025),so fail to reject the null hypothesis.

No, change in conclusion.  

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Nadya [2.5K]

Answer:

The probability that the average score of the 49 golfers exceeded 62 is 0.3897

Step-by-step explanation:

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5

\mu = 61

\sigma = 3.5

We are supposed to find he probability that the average score of the 49 golfers exceeded 62.

Formula : Z=\frac{x-\mu}{\sigma}

Z=\frac{62-61}{3.5}

Z=0.285

Refer the z table for p value

p value = 0.6103

P(x>62)=1-P(x<62)=1-0.6103=0.3897

Hence the probability that the average score of the 49 golfers exceeded 62 is 0.3897

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Answer:

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Step-by-step explanation:

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