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/2/squareroot n and Confidence Interval mu plusminus E with E = t* middot s/squareroot n 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: Juveniles: 65513 65934 64213 61914 59167 Adults: 31364 34478 36937 35946 38209 Test the claim at 0.05 level of significance. Test statistic t = mu_1 - mu_2/squareroot s^2_2/n_2 + s^2_2/n_1 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. b. 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 cap plusminus z_x/z squareroot pq/n

Explanation / Answer

hypothesis

H0: average days=10

H1: average daysnot equal to 10

mean=8.375

std deviation=4.1036

test statistics=t=(8.375-10)/(4.1036/sqrt8)=-1.120039496

The P-Value is .299665

as p value>0.05

conclusion

we fail to reject the Null hypothesis and conclude that there is not enough evidence to support the claim that The average number of sick days an employee takes per year is not equal to 10 days

confidence intervel

X+/- E

Critical value of t for 5% significance level and degrees of freedom (n - 1 = 7) is +2.364

here E=t*s/sqrt(n)

=3.4298

C.I = [8.375-3.4298 ,8.375+3.4298]

=(4.9452,11.8048)

0 is between confidence intervel , we can say that the result is not significant at 0.05

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