Question

In: Statistics and Probability

A company claims that 9 out of 10 doctors (i.e. 90%) recommend its brand of cough...

A company claims that 9 out of 10 doctors (i.e. 90%) recommend its brand of cough syrup to their patients. To test this claim against the alternative that the actual proportion is less than 90%, a random sample of 200 doctors was chosen which results in 160 who indicate that they recommend this cough syrup. Find the standardized test statistic, z. Round to two decimal places.

Expert Solution

Let's choose level of significance = 5% ( since it is not given )

We have our hypothesis defined as :

Null hypothesis H0: p >= 0.90 i.e actual proportion is great than or equal to 0.9

Alternative hypothesis H1: p < 0.90 i.e. actual proportion is less than 0.9

Therefore, this is a left tailed left.

The one proportion z- test is given by :

where,

z = Test statistics
n = Sample size = 200
po = Null hypothesized value = 0.9
= Observed proportion = 160/200 = 0.8

z = = = - 4.716

The z- score at 5% level of significance for left tailed test is calculated as follows :

1. Calculate 0.5 - 0.05 = 0.45 ( 0.05 is the level of significance )

2. Spot value closest to 0.45 in the z- table.

3. Look for the column and row label, put the column label at the second decimal point of row label and put minus sign.

here, Row label = 1.6 and column label = 0.05

So, we have our z - score = - 1.65

Now, since our Z_calculated < Z _critical i.e -4.716 < -1.65 , we reject out null hypothesis and accept the alternate hypothesis.

Therefore, we accept the alternate hypothesis that the actual proportion is less than 0.9

orchestra answered 2 years ago

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