The mean number of sick days an employee takes per year is believed to be about

Question

The mean number of sick days an employee takes per year is believed to be about 10-Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Should the personnel team believe that the mean number is about 10? Test statistic t = x - mu/5/squareroot n and Confidence Interval Mu plusminus E with E = t + 5/squareroot n State the null and alternative hypothesis. State your conclusion for your hypothesis test. Give the 95% confidence Interval. Give a conclusion for the confidence interval test. A researcher wants to test the claim that, on average, more juveniles than adults are classified as missing persons. Records for the last 5 years were recorded as follows: Test the claim at 0.05 level of significance. Test statistic t = Mu_1 - Mu_2/squareroot s^2_1/n_1 - s^2_2/n_1 State the null and alternative hypothesis. State your conclusion for your hypothesis test. A new insectercide is advertized to kill more than 95% of roaches on contact. The insectercide was applied to 400 roaches. Only 384 died immediately after contact. a. Is this sufficient evidence to support their manufacturer's claim of 95%. Use alpha = 0.05 Level of significance. Find a 95% confidence interval for the true proportion of roaches killed on contact. Test statistic z = p^- p^_0/squareroot p_0 - (1 - p_0)/n and Confidence Interval p^plusminus z_alpha/2 squareroot p^q^/n State the null and alternative hypothesis. State your conclusion for your hypothesis test. Give a conclusion for the confidence interval test.

Explanation / Answer

1)

hypothesis:

H0: mu = 10
Ha: mu not equal to 10

Data:
mean = 10 , x = 8.375 , s = 4.106, n = 8

Tests statistic:

t = ( x - mean) / (s/sqrt(n))
= ( 8.375 - 10)/ ( 4.106 / sqrt(8))
= -1.119

p value is calculated using t = -1.119 , df = 7
P value = .300063.

we fail to reject the null hypothesis.

t value for 95% CI = 2.365

CI = mean + /- t * ( s / sqrt(n))
= 10 + /- 2.365 * (4.106 / sqrt(8))
= (6.567 , 13.433)

3)

3)
Null hypothesis: P <= 0.95
Alternative hypothesis: P > 0.95

Data:
p = 384 / 400 = 0.96
= sqrt[ P * ( 1 - P ) / n ]
= sqrt [(0.95 * 0.5) / 400]
= 0.011
z = (p - P) /
= (.96 - .95)/0.011
= 0.909

p value is calculated using z = 0.909
P value = 0.181675.

Here, we fail to reject the null hypothesis.

z value at 95% CI = 1.96

CI = p +/- z * sqrt(pq / n)
= 0.96 +/- 1.96 * sqrt( 0.96 * 0.04 / 400)
= (0.9407 , 0.9792)

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