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Klio2033 [76]
3 years ago
10

Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come f

rom a random sample and use a 5% significance level.
Test H0: p = 0.25 vs Ha: p < 0.25 using the sample results p(hat) = 0.15 with n = 102
Round your answer for the test statistic to two decimal places, and your answer for the p-value to three decimal places.
Mathematics
1 answer:
sveta [45]3 years ago
7 0

Answer:

Test statistic = - 2.33

Pvalue = 0.010 (3 decimal places)

Step-by-step explanation:

Given :

H0: p = 0.25 vs Ha: p < 0.25

Phat = 0.15

Sample size, n = 102

α = 0.05

The test statistic :

(Phat - p) / (√p(1 - p) /n)

1 - p = 1 - 0.25 = 0.75

(0.15 - 0.25) / (√0.25(0.75) /102)

-0.1 / √0.0018382

-0.1 / 0.0428742

Test statistic = - 2.33

The Pvalue using the test statistic score obtained :

P(Z < - 2.33) = 0.0099031 = 0.010

Pvalue < α ; Reject H0

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3 years ago
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The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
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Answer:

(a) 11.25 and 1.68  

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 15

             r = number of success

             p = probability of success which in our question is % of cases in

                  which customer problems are resolved on the same day, i.e.;75%

So, here X ~ Binom(n=15,p=0.75)

(a) Expected number of problems to be resolved today = E(X)

            E(X) = \mu = n * p = 15 * 0.75 = 11.25

    Standard deviation = \sigma = \sqrt{n*p*(1-p)} = \sqrt{15*0.75*(1-0.75)} = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

     P(X = 10) = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}

                    = 3003*0.75^{10} *0.25^{5} = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

    = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}

    = 3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4} = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

    given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)  

= \binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}

= 1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}

= 0.6865

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