Researchers studied the friction that occurs in the Paper-feeding process of a p

Question

Researchers studied the friction that occurs in the Paper-feeding process of a photocopier. In a stack of 100 sheets of paper, the feeding process was successful 94 times. The success rate of the feeder is designed to be .90. Test to determine if the success rate of the feeder exceeds .90 at alpha = 0.05. Test statistic Z = p - p_0/squareroot p_0 (1 - p_0)/n and Confidence Interval p plusminus Z_alpha/2 squareroot pq/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. As a professional courtesy, physicians have traditionally provided health care free of charge or at a reduced rate to other physicians and their families. In 1986, 94% of a sample of 1,000 physicians offered this professional courtesy. To assess the extent to which this practice has changed the years, The New England Journal of Medicine conducted a survey of 2, 224 physicians of which 1, 957 currently offer free or reduced-rate health care to fellow physicians. Test statistic Z = beta_1 - beta_2/squareroot beta q (1/n_1 + 1/n_2) with p = x_1 + x_2/n_1 + n_2 State the null and alternative hypothesis. State your conclusion for your hypothesis test. In a study of drug usage, researchers surveyed the type of drug fist injected by 102 subjects with results as follows: Heroin 42 Speed 36 Other 24 At alpha = .05, test that the probabilities for the three groups were equal. (1/3, 1/3, 1/3). Chi Squared Formula: X^2 = sigma^n _i=1 (O_i - E_i)^2/E_i State the null and alternative hypothesis. State the observed values. State the expected values. Give a conclusion for the confidence interval test.

Explanation / Answer

Question 3)

State the null and alternative hypothesis:

H0: P 0.90

H1: P > 0.90

State your conclusion for your hypothesis test:

Z = (p^ - P)/sqrt((P*(1-P))/n)

   = ((94/100)-0.90)/sqrt((0.90*(1-0.90))/100)

= 1.33

From the table we get the critical z as 1.645.

The test statistics is less than the critical value; we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the success rate of the feeder exceeds .90 at 5% level of significance.

Give the 95% confidence interval:

P^ (-/+) E

p^ = 94/100 = 0.94

q^ = 1 – p^

     = 1 -0.94

E = zc * sqrt ((p^*q^)/n)

From critical table we get the critical zc value for two-tailed as 1.96

E = 1.96*sqrt((0.94*(1-0.94))/100)

P^ (-/+) E

0.94 (-/+) 0.0465

0.893 and 0.987

The 95% confidence interval is (0.893, 0.987)

Give a conclusion for the confidence interval test:

We are 95% confident that the population proportion will lie in between 0.893 and 0.987.

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