Statistics assignment 3 PDF

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Statistics assignment 3...

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QUESTION 1: a)

For samples of size n = 500, the sampling distribution of ^p has a mean of  ^p = p = 0.75 and a standard error of  ^p = p(1− p) n 0.25 ¿ p (1− p ) SE ^p = = 0.75 ¿ = 0.01936. n ¿ √¿ When n is large, the distribution of ^p is approximately normal.

√

b)

As the sample of size n = 33 is obtained from a normally distributed population the Central Limit Theorem has little work to do. This is because the population measurements are already approximately normally distributed. Therefore, we can say that sample means will be normally distributed for samples of n = 33 from this population.

QUESTION 2: a) i) H0: p = 0.66667 H1: p > 0.66667 where p is the proportion of adult Australian internet users that get health information from the internet. ii)

significance level -  = 0.01

iii) Since the sample size is large and is randomly sampled the sampling distribution of ^p will follow an approximate normal distribution. iv)

v)

1387 x = = 0.69909 1984 n test statistic is Z = ^p – p0  √ p 0 (1− p 0)/n = 0.69909 – 0.66667  √ 0.66667 x 0.33333 / 1984 = 0.03242  0.010583 = 3.06331 = 3.06 ^p =

p-value = P ( Z > 3.06) = 1 – P( Z < 3.06) = 1 – 0.9989 = 0.0011

vi) Since the p-value (0.0011) < 0.01, we reject H0 at 1% significance level. The small p-value indicates that there is practically no chance that the observed ^p could arise from the sampling variation if H0 were true.

It appears that 69% -70% of Australian adult internet users get health information from the internet. b) As the sampling method that was used to obtain the sample from the population was simple random sampling we can assume the conclusion from part a could be generalised to the Australian adult population. Simple random sampling is the “gold standard” of sampling and it states that every “case” in the population has the same chance of being included in the sample. QUESTION 3: a) It would not be reasonable to treat “a lab class of 10 biochemistry students” as a representative sample of wine consumers as the sample is not randomly selected. Students are very unlikely to be “trained” at wine tasting and as the sample is not selected randomly there is likely to be bias present. Convenience sampling has been used in this sample selection leading to bias. Also the size of the sample is very small and therefore may not be a good representation of the population as a whole. b)

One-Sample Statistics N DMSthresho ld

10

Mean 30.40

Std. Deviation 6.753

Std. Error Mean 2.135

One-Sample Test Test Value = 25

DMSthresho ld

t 2.529

df 9

90% Confidence Interval of the Difference Sig. (2Mean Lower Upper tailed) Difference .032 5.400 1.49 9.31

90% CI for  = 25 = 1.49 + 25 to 9.31 + 25 = 26.49 g/L to 34.31 g/L c) i) H0 :  = 25 vs H1 :   25 where  is the mean dimethyl sulphide (DMS) odour threshold among wine consumers ii)

P-value = 0.032

iii) since p-value (0.032) is <  = 0.05, we reject H0 at 5% significance level. Therefore, the mean DMS odour threshold is greater than 25 g/L (micrograms per litre). The estimated mean DMS odour threshold for the lab class of 10 biochemistry students is 30.40 g/L.

QUESTION 4: z 2 ) p(1-p) e 1.645 2 ) x 0.08(0.92) = 1991.63  1992 sample size =n( 0.01

n(

QUESTION 5: The data (i.e., the aspirin concentrations) should be analysed as paired data. In this investigation the researchers took pairs of measurements (the subjects took either aspirin a or b first and then later took the other aspirin that they had not been given initially, so that each subject took both aspirins in the trial) and then difference between each pair was calculated and was analysed. This is a classic matched pairs design. QUESTION 6: a)

i) H0 = c = t H1 =  c   t where  is the wound healing rate in newts in this population and where subscripts c & t indicate treatment group and control group respectively. b)

ii) There was no simple random sampling from the population of all newts. It is assumed that the set of animals is an SRS from their species if there was nothing special about how the subjects were obtained. We have an independent variable that is categorical and the samples are randomly selected (i.e. 33 newts) - CLT ensures that the difference in the sample means will be approximately normally distributed. Distribution is roughly symmetric with no outliers and therefore all necessary conditions are met. iii)

significance level -  = 0.05

Group Statistics EField HealDiff Control group Treatment group

N

Std. Deviation 9.345 17.387

Mean -.17 -13.80

18 15

Std. Error Mean 2.203 4.489

Independent Samples Test

Levene's Test for Equality of Variances

t-test for Equality of Means 95% Confidence Interval

F HealDif Equal variances f

3.765

Sig.

t .061

df

Sig. (2-

Mean

Std. Error

tailed)

Difference

Difference

of the Difference Lower

Upper

2.872

31

.007

13.633

4.748

3.951

23.316

2.726

20.570

.013

13.633

5.001

3.221

24.046

assumed Equal variances not assumed

iv)

p-value = 0.013

v)

Test statistic = t = ´x c – ´x t /√(sc2 /nc) + (st2/nt) = -0.17- (-13.80) / √(9.3452/18) + (17.3872/15) = 13.63/ √4.8516 + 20.1538 = 13.63/√25.0054 = 13.63/5.00054 = 2.726

Since the p-value 0.013 is <  0.05 we reject H0 at 5% significance level. There is strong evidence that a 50% increase in the natural electric field would have no effect on the mean wound healing rates in newts. The difference in wound healing rates is 13.633 m/h....