*Need help with part (a) and (c) Heres the problem: GreenBeam Ltd. claims that i

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*Need help with part (a) and (c)

Heres the problem:

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.29 mg of mercury. A sample of 35 bulbs shows a mean of 3.38 mg of mercury.

Assuming a known standard deviation of 0.2 mg, calculate the z test statistic to test the manufacturer’s claim. (Round your answer to 2 decimal places.)

  

Find the p-value. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.)

  

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.29 mg of mercury. A sample of 35 bulbs shows a mean of 3.38 mg of mercury.

(a)

Assuming a known standard deviation of 0.2 mg, calculate the z test statistic to test the manufacturer’s claim. (Round your answer to 2 decimal places.)

  

(b) At the 1 percent level of significance ( = 0.01) does the sample exceed the manufacturer’s claim? Yes (c)

Find the p-value. (Round intermediate calculations to 2 decimal places. Round your answer to 4 decimal places.)

  

  

Explanation / Answer

a). Consider the given problem here the sample size is “35” sample mean is “3.38” and known population mean is “0.2”.

So, given the information the null hypothesis is “H0: µ=3.29” and the corresponding alternative hypothesis is “H1: µ > 3.29”. So, under “H0”, the calculated value of the “z” statistic is, “n*(sample mean – 3.29)/sd”

=> z = n*(sample mean – 3.29)/sd = 35*(3.38 – 3.29)/0.2 = 0.5324/0.2 = 2.66.

So, given the information in the given question the value of “z” is “2.66”.

b). This is a one sided test, where “z” is “2.66” and at “1%” level of significance the critical value is “2.33 < 2.66”, => the “z” is on the critical region, => we will reject “H0” and will conclude that since the sample mean is significantly different from the”3.29” given the information mention in the problem.

c). As we know that “z” follows “standard normal” distribution, and “z=2.66”, => the “p” value of “z” is nothing but the right sided area under the standard normal distribution of the “2.66”, since "H1: µ > 3.29. So, the area under the standard normal distribution of “z=2.66” is “1-0.99609 = 0.0039 = 0.39% < 1% (by using standard normal table).

So, the “p” value of “z” is “0.39%”. So, since “p” value is < 1%, => if we set “level of significance at 1%”, => p < level of significance, => “H0” will be rejected and “H1” will be accepted.

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