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Morgarella [4.7K]
3 years ago
11

A paint company produces glow in the dark paint with an advertised glow time of 15 min. A painter is interested in finding out i

f the product behaves worse than advertised. She sets up her hypothesis statements as H0 : µ ≤ 15 and Ha : µ > 15, then calculates a test statistic of z = −2.30. What would be the conclusions of her hypothesis test at significance levels of α = 0.05, α = 0.01, and α = 0.001?
Engineering
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

p_v = P(z

Now we can decide based on the significance level \alpha. If p_v we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

\alpha=0.05 we see that p_v< \alpha so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

\alpha=0.01 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

\alpha=0.001 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

Explanation:

For this case they conduct the following system of hypothesis for the ture mean of interest:

Null hypothesis: \mu \leq 15

Alternative hypothesis: \mu >15

The statistic for this hypothesis is:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And on this case the value is given z = -2.30

For this case in order to take a decision based on the significance level we need to calculate the p value first.

Since we have a lower tailed test the p value would be:

p_v = P(z

Now we can decide based on the significance level \alpha. If p_v we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

\alpha=0.05 we see that p_v< \alpha so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

\alpha=0.01 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

\alpha=0.001 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

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Answer:

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at mid span

\delta = \frac{wl^3}{48EI}

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\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

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