Use 0.10 as the significance level (α).
Conduct a one-sample hypothesis test to determine if the networks should announce at 8:01 P.M. the Republican candidate George W. Bush will win the state.
In order to conduct a one-sample hypothesis test, the first step is to develop the hypotheses.
Our hypotheses in this case will be;
H 0 : p=0.5 (null hypothesis)
H 1 : p>0.5 (alternative hypothesis)
To test the claim formulated above the formula to be used is:
z = where is the sample proportion
is the hypothesized proportion
is 1-
n is the sample size
In this case, the above formula is applied because it deals with a population proportion i.e. what proportion of registered voters in Florida state voted for the democratic presidential candidate Al Gore as compared to the republican presidential candidate George Bush.
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Given n=765
Gore’s vote= 358
Bush’s votes= 407
Therefore;
= 407/765=0.532026
Hence;
z = = 1.771596
the rejection region from the z table value at 0.1 is 2.578
z cal (1.771596) < z tab (2.578)
Decision: do not reject H 0
Conclusion: George Bush, the republican presidential candidate, will fail to win the state therefore the network should not announce him as the winner.
Analysis
The exit polls done by the network companies interviewed a total number of 765 voters exiting polling booths. Since the majority of those interviewed stated they voted for the republican candidate, George Bush (407) as compared to 358 voters who voted for Al Gore, the notion is that President Bush will win this state. The network station however announces a winner only if they garnered over 50% of the total votes cast.
To test this claim that George Bush wins in Florida state a hypothesis is developed and the null hypothesis will be H 0 : p=0.5. This is because mean and standard deviations are not known, the central limit theorem is used while making the assumption that the population distribution is normal. The alternative hypothesis will be that the republican candidate will win the election by getting more than 50% will be H 1 : p>0.5.
After calculating the z value 1.771596 is obtained and this has to be compared to the tabulated value at 0.1α level of significance which is 2.58. If the null hypothesis is found not to be plausible, it is rejected otherwise, it is accepted. In this case, the calculated value is less than the value from the z table and this implies that the null hypothesis is reasonable therefore can be accepted.
Accepting the null hypothesis means that the republican presidential candidate George Bush does not get more than 50% of the total votes cast therefore the network should not declare him as the winner at 8:01pm. The win margin is not significant enough.
Case 2: SpeedX
Use 0.10 and the significance level (α).
Conduct a one-sample hypothesis test and determine if you can convince the CFO to conclude the plan will be profitable.
To conduct a one-sample hypothesis test, the first step is to develop the hypotheses (Scheaffer, Mendenhall, & Ott, 1995) .
Our hypotheses are;
H 0 : µ=24 days (null hypothesis)
H 1 : µ<24days (alternative hypothesis)
To test the claim the formula to used is:
z = where is the sample mean
is the hypothesized mean
n is the sample size
Given mean (µ) to be 24, standard deviation (sigma) as 6 and sample size 220
Calculating the claim;
z = = -5.85434
the z tabulated value at 0.1α level of significance is 2.578
therefore;
z cal (-5.85434) < z tab (2.578)
Decision: do not reject H 0
Conclusion: the plan is not profitable
Analysis
In this study, the chief financial officer (CFO) of Speedx courier company believes that having envelopes which are self-stamped would help decrease the amount of time taken by consumers to pay for the purchased goods as currently it takes customers an average of 24 days to pay up with a standard deviation of 6 days. To test this claim the hypotheses have to be formulated. The null hypothesis which is the hypothesis of no difference would be that the number of days taken to pay for goods does not decrease even after including a stamped self-addressed (which is 24 days on average) H 0 : µ=24 days against the alternative hypothesis that including a stamped self-addressed envelope does decrease the number of days of paying for goods H 1 : µ<24days (Inglehart, 2000) .
In testing this claim the calculated z value is -5.85434 and this has to be compared to the tabulated value at 0.1α level of significance which is 2.58. If the null hypothesis is found not to be plausible, it is rejected otherwise, it is accepted. In this case, the calculated value is less than the value from the z table and this implies that the null hypothesis is reasonable therefore can be accepted.
This implies that even if the courier company decides to include stamped self-addressed envelopes, there will not be a significant drop in the number of days it takes customers to pay them. From the pilot study conducted the mean number of days it takes for the company to get paid while using this plan is approximately 22 days and the one-sample hypothesis test proves that this decrease in the number of days is not significant enough.
Therefore, I would advise the CFO not to use this plan as it is not profitable as the number of days it will take customers to pay will still be high.
References
Inglehart, W. (2000). Hypothesis (1st ed.). New York, N.Y.: Vantage Press.
Scheaffer, R., Mendenhall, W., & Ott, L. (1995). Elementary survey sampling (1st ed.). Belmont: Duxbury Press.
