Statisticsassignmentexperts.com is an organisation committed to providing world class education solutions in the subject of Statistics to students across the globe. We specialise in offering a multitude of educational services to students for Statistics, who are studying in various institutes and universities, be it assignments, online tutoring, project work, dissertation/thesis, or exam preparation. When it comes to information and communication technologies, we employ state-of-the-art and latest tools and technology to connect with students and expert tutors. Students from various countries including the USA, UK, Canada, UAE and Australia have used our services for the past several years to achieve excellence in their academic and professional pursuits. Statisticsassignmentexperts.com works closely with its strong and dynamic team of subject experts to create new models for exchange of information, in consonance with the changing needs of students as well as academic and professional programs.

1. Testing Of Hypothesis
1. Iowa GPA:
0H : μ= 2.80
aH : μ≠2.80
In the above hypotheses, 0H : is the notation for the null hypothesis, aH : is the notation for the
alternative hypothesis, and μ is the parameter, the mean GPA of the population, about which
we’re hypothesizing.
2. P-value:
a) This P-value does not give strong evidence against the null hypothesis.
b) This extreme P-value does give strong evidence against the null hypothesis.
3. Another test of therapeutic touch:
a) p = proportion of trials guessed correctly
0H : p = 0.50 and aH : p > 0.50
b) pˆ = 53/130 = 0.4077
se =    npp /1    130/50.150. 0.0439
z = (0.4077 – 0.5)/0.0439 = - 2.10; the sample proportion is a bit more than 2 standard errors
less than would be expected if the null hypothesis were true.
c) The P-value is 0.9821 (rounds to 0.98). Because the P-value is so much greater than the
significance level of 0.05, we do not reject the null hypothesis. The probability would be 0.98
of getting a test statistic at least as extreme as the value observed if the null hypothesis were
true, and the population proportion were 0.50.
d) np = (130)(0.5) = 65= n(1-p); the sample size was large enough to make the inference in (c).
We also would need to assume randomly selected practitioners and subjects for this to apply
to all practitioners and subjects.
4. Garlic to repel ticks:
a) The relevant variable is whether garlic or placebo is more effective, and the parameter is the
population proportion, p = those for whom garlic is more effective than placebo.
b) 0H : p = 0.50 and aH : p ≠ 0.50; the sample size is adequate because there are at least 15
successes (garlic more effective) and failures (placebo more effective).
c) pˆ = 37/66 = 0.561
se =    npp /1    66/50.150. 0.062
z = (0.561 – 0.5)/0.062 = 0.984
d) The P-value is 0.325 (rounds to 0.33). This P-value is not that extreme. If the null hypothesis
were true, the probability would be 0.33 of getting a test statistic at least as extreme as the
value observed. It is plausible that the null hypothesis is correct. Because the probability is
0.33 that we would observe our test statistic or one more extreme due to random variation,
there is insufficient evidence that the population proportion which would have fewer tick bites
with garlic vs. placebo is not different from 0.50.
5. Which cola?:
a) The test statistic (“Z-Value”) is calculated by taking the difference between the sample
proportion and the null proportion and dividing it by the standard error.
b) We get the “P-value” by looking up the “Z-value” in Table A or using software. We have to
determine the two-tail probability from the standard normal distribution below -1.286 and

2. above 1.286. The P-value of 0.20 tells us that if the null hypothesis were true, a proportion of
0.1985 (rounds to 0.20) of samples would fall at least this far from the null hypothesis
proportion of 0.50. This is not very extreme; it is plausible that the null hypothesis is correct,
and that Coke is not preferred to Pepsi.
c) It does not make sense to accept the null hypothesis. It is possible that there is a real
difference in the population that we are not detecting in our test, and we can never accept a
null hypothesis.
d) The 95% confidence interval tells us the range of plausible values, whereas the test merely
tells us that 0.50 is plausible.
6. Lake pollution:
a) From software, the mean of the four observations is 2000; the standard deviation is 816.5;
and the standard error is 408.25.
b) Software gives a t-score of 2.45.
c) The P-value is 0.046 for a one-sided test. This is smaller than 0.05, so we have enough
evidence to reject the null hypothesis at a significance level of 0.05. We have strong evidence
that the wastewater limit is begin exceeded.
d) The one-sided analysis in (b) implicitly tests the broader null hypothesis that μ≤1000. We
know this because if it would be unusual to get a sample mean of 2000 if the population
mean were 1000, we know that it would be even more unusual to get this sample mean if the
population mean were less than 1000.
7. Selling a burger:
1) Assumptions: The data are quantitative (dollar differences). The data might have been
produced randomly.
2) Hypotheses: 0H : μ = 0; aH : μ ≠ 0
3) Test statistic: 


10/4000
03000
t 2.372
4) P-value: 0.04177 (rounds to 0.042)
5) Conclusion: If the null hypothesis were true, the probability would be 0.042 of getting a test
statistic at least as extreme as the value observed. Because the P-value is 0.04, which is <
0.05, there is sufficient evidence that the coupons led to higher sales than did the outside
posters.
8. Decision errors in medical diagnostic testing:
a) A Type I error is a false positive because we have rejected the null hypothesis that there is no
disease, but we were wrong. The woman in fact does not have breast cancer. The
consequence would be that the woman would have treatment, or at least further testing,
when she did not need any.
b) A Type II error is a false negative because we have failed to reject the null hypothesis that
there is no disease, but we were wrong. The woman in fact does have breast cancer. The
consequence would be failing to detect cancer and treat the cancer when it actually exists.
c) The disadvantage of this tactic is that more women who do have breast cancer will have false
negative tests and not receive necessary treatment.
9. Misleading summaries?:
a) Researcher A:
se =    npp /1    400/50.150. 0.025
z = (0.55 – 0.50)/0.025 = 2.00
P-value =2P(z >2.0) =0.046. .
b) Researcher B:
z = (0.5475 – 0.50)/0.025 = 1.90

3. P-value =2P(z >1.90)= 0.057.
c) Researcher A’s result has a P-value less than 0.05; thus, it is “statistically significant.”
Researcher B’s P-value is not less than 0.05, and is not, therefore, “statistically significant.”
Results that are not different from one another in practical terms might lead to different
conclusions if based on statistical significance alone.
d) If we do not see these two P-values, but merely know that one is statistically significant and
one is not, we are not able to see that the P-values are so similar.
e) We can add and subtract the result of (1.96)(0.025) – the z-score associated with a 95%
confidence interval multiplied by the standard error – to the mean for each of these samples
to get confidence intervals of (0.501, 0.599) for Researcher A and (0.499, 0.596) for
Researcher B. This method shows the enormous amount of overlap between the two
confidence intervals. The plausible values for the population proportions are very similar in
the two cases, which we would not realize by merely reporting whether the null was rejected
in a test.