solve the following problems. follow the steps in hypothesis testing.?
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1.a professor find the class an examination that, as he knows from years of experience, yields μ=78 and σ=7. the present class of 22 obtains a mean of 82. Is the professor, correct in assuming that this is a superior class? Use α=0.01, two tailed test.
2. a company makes chocolates claims that the mean weight of a bag of chocolates is 240 grams with a standard deviation of 20.56 grams. Using 5% significance level, would you agree with the company if a random sample of 50 bags of chocolate was found to have a mean weight of 230 grams?
3. managers of a diet plant advertise that the mean weight loss for people on thier plan is at least 45 pounds in 6 months. A sample of 28 people on this plan loss an average of 35 pounds with a standard deviation of 20 pounds. test, at 1% level of significance if the claim is high.
oh, i know this.i just don’t have my notes with me now.
for theses three questions we will be using a hypothesis test for the mean.
Hypothesis Test for mean:
assuming you have a large enough sample such that the central limit theorem holds and the mean is normally distributed then to test the null hypothesis H0: μ = Δ
find the test statistic z = (xBar – Δ) / (sx / sqrt(n))
where xbar is the sample average
sx is the sample standard deviation
n is the sample size
The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.
H1: μ > Δ; p-value is the area to the right of z
H1: μ < Δ; p-value is the area to the left of z
H1: μ ≠ Δ; p-value is the area in the tails greater than |z|
for a small sample test for the mean every thing is the same save the test statistic is a t statistic with n - 1 degrees of freedom. However, in this case the underlying distribution must be normal for this test to be valid.
1. Although n = 22 is a small sample size, the fact that we know the population standard deviation σ=7 means we can use the z test.
the hypothesis test we are looking at is:
H0: μ ≤ 78 vs H1: μ > 78
The test statistic is z = (82 – 78) / (7 / √(22) ) = 2.680238
the p-value of the test is:
p-value = P(Z > 2.680238) = 0.003678492
since the p-value is less than the significance level we will reject the null hypothesis and conclude the alternate hypothesis is true, this class has a mean greater than 78.
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2.
H0: μ = 240 vs. H1: μ ≠ 240
the test statistic is: z = (230 – 240) / (20.56 / sqrt(50) )
z = -3.439235
the p-value is:
p-value = P( |Z| > 3.439235)
= P( Z < -3.439235 ) + P( Z > 3.439235)
= 2 * P( Z < -3.439235)
= 2 * 0.0002916803
= 0.0005833606
since the p-value is less than the significance level we will reject the null hypothesis and conclude the alternate is true, the bags do not weigh 240 grams. Note that since we did a two tail test we cannot conclude the bags weigh less than 240 grams, just that the do not weight 240 grams.
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3.
H0: μ ≥ 45 vs H1: μ < 45
the test statistic is going to be a t-statistic here because we have a small sample size and a sample standard deviation. In truth, any time you have a sample standard deviation the t test will be better to use, regardless of sample size. However, over 30 degrees of freedom the quantiles of the student t are very close to the standard normal which is why there is the "30 samples to assume normality" assumption.
t = (35 - 45) / (20 / sqrt(28)) = -2.645751
the p-value is going to be the area under the student t with 27 degrees of freedom less than the test statistic
p-value = P( t_27 < -2.645751) = 0.006712699
since the p-value is less than the significance level we will reject the null and conclude that the mean weight loss is less than 45 pounds.
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If you have a difficult time choosing the direction of the inequalities for a hypothesis test then the following should help.
Consider the hypothesis as a trial against the null hypothesis. the data is evidence against the mean. you assume the mean is true and try to prove that it is not true.
If the question statement asks you to determine if there is a difference between the statistic and a value, then you have a two tail test, the null hypothesis, for example, would be μ = d vs the alternate hypothesis μ ≠ d
if the question ask to test for an inequality you make sure that your results will be worth while. for example. say you have a steel bar that will be used in a construction project. if the bar can support a load of 100,000 psi then you'll use the bar, if it cannot then you will not use the bar.
if the null was μ ≥ 100,000 vs the alternate μ < 100,000 then will will have a meaningless test. in this case if you reject the null hypothesis you will conclude that the alternate hypothesis is true and the mean load the bar can support is less than 100,000 psi and you will not be able to use the bar. However, if you fail to reject the null then you will conclude it is plausible the mean is greater than or equal to 100,000. You cannot ever conclude that the null is true. as a result you should not use the bar because you do not have proof that the mean strength is high enough.
if the null was μ ≤ 100,000 vs. the alternate μ > 100,000 and you reject the null then you conclude the alternate is true and the bar is strong enough; if you fail to reject it is plausible the bar is not strong enough, so you don’t use it. in this case you have a meaningful result.
Any time you are defining the hypothesis test you need to consider whether or not the results will be meaningful.