We are given that a manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly. The population standard deviation is 50 hours and the light bulb life is normally distributed.

You select a sample of 100 light bulbs and find mean bulb life is 490 hours.

Let $\mu$ = population mean light bulb life.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 500 hours {means that the population mean light bulb life is at least 500 hours}

Alternate Hypothesis, $H_A$ : $\mu$ < 500 hours {means that the population mean light bulb life is below 500 hours}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean bulb life = 490 hours

σ = population standard deviation = 50 hours

n = sample of light bulbs = 100

So, the test statistics = $\frac{490-500}{\frac{50}{\sqrt{100} } }$

= -2

The value of z test statistics is -2.

Now, at 0.01 significance level the z table gives critical value of -2.33 for left-tailed test.

Since our test statistic is higher than the critical value of z as -2 > -2.33, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population mean light bulb life is at least 500 hours.

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3 years ago

You have been saving $25 a week to make the trip; you now have $200. This money will be budgeted to include gas, food, and enter

Pavel [41]

Domain is 25 & 200
Range is 175

3 0

1 year ago