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A hospital director believes that 45% of the lab reports contain errors. A sample of 260...

A hospital director believes that 45% of the lab reports contain errors. A sample of 260 reports found 104errors. Is there sufficient evidence at the 0.10 level to refute the hospital director's claim?

State the null and alternative hypotheses for the above scenario.

Math   Statistics-And-Probability   
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Answer #1

Answer:

This is the two tailed test .

The null and alternative hypothesis is

H0 : p =0.45

Ha : p 0.45

= x / n = 104/260

=0.4

1 - P0 = 1-0.45

=0.55

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

=0.4-0.45 / [(0.45*0.55) / 260]

= -1.69

P(z <-1.69 ) = 0.0455

P-value = 2*0.0455=0.091

= 0.10

P-value <

Reject the null hypothesis .

There is sufficient evidence to suggest that  

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