A publisher reports that 79% of their readers own

Ответ:

marieb39

22.01.2022

The value of the test statistic is -1.91.

Step-by-step explanation:

We are given that a publisher reports that 79% of their readers own a laptop. A random sample of 150 found that 72% of the readers owned a laptop.

A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.

Let p = population % of readers who own a laptop

SO, Null Hypothesis, H_0 : p = 79% {means that the percentage is same as that of the reported percentage}

Alternate Hypothesis, H_a : p $\neq$ 79% {means that the percentage is actually different from the reported percentage}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = % of the readers who owned a laptop in a sample of 150 readers = 72%

n = sample of readers = 150

So, test statistics = $\frac{0.72-0.79}{\sqrt{\frac{0.72(1- 0.72)}{150} } }$

= -1.91

Therefore, the value of the test statistic is -1.91.

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