mathcentre: Parametric hypothesis testing for psychology

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Marks Available:	17
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Question 1

The Royal medical group makes the following claim:

The average cost of a treatment of benzodiazepines with us is just £71 (including all taxes and charges!)

A rival medical company decides to test their claim.

A sample of $\var {n}$ patients is taken.

The mean cost of treatment from this sample is £ $\var {m}$ with a standard deviation of £ $\var{stand}$ .

Perform an appropriate hypothesis test to see if the claim made by the Royal medical group is substantiated (use a two-tailed test).

Advice

a)

Step 1: Null Hypothesis

$\operatorname{H}_0\;: \; \mu=\;\var{thisamount}$

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;: \; \mu \neq\;\var{thisamount}$

b)

We should use the $t$ statistic as the population variance is unknown.

The test statistic:

$t =\frac{ |\var{m} -\var{thisamount}|} {\sqrt{\frac{\var{stand} ^ 2 }{\var{n}}}} = \var{tval}$

to 3 decimal places.

c)

As $n=\var{n}$ we use the $t_{\var{n-1}}$ tables. We have the following data from the tables:

p value	10%	5%	1%
Critical Value	1.706	2.056	2.779

We see that the $p$ value lies between 1% and 5%.

d)

Hence there is moderate evidence against $\operatorname{H}_0$ and so we reject $\operatorname{H}_0$ .

There is sufficient evidence against the claim of the Royal medical group.

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Question 2

A phone company wants to know if there is any difference between the average time spent on the telephone, between cat lovers and dog lovers.

A random sample of $\var{n1}$ cat lovers and $\var{n2}$ dog lovers gave the following results in seconds.

	Mean	Standard deviation
Cat lovers	240	60
Dog lovers	360	82

Perform an appropriate hypothesis test to see if there is any difference in the average time spent on the telephone between cat lovers and dog lovers (the null and alternative hypotheses have been set out for you).

Advice

b)

Step 3 : Test statistic

We should use the $t$ statistic as the population variance is unknown.

The pooled standard deviation is given by :

\[s = \sqrt{\frac{\var{n1 -1} \times \var{sd} ^ 2 + \var{n2 -1} \times \var{sd1} ^ 2 }{\var{n1} + \var{n2} -2}} = \var{tpsd} = \var{psd}\] to 3 decimal places.

The test statistic is given by \[t = \frac{|\var{m} -\var{m1}|}{s \sqrt{\frac{1}{ \var{n1} }+\frac{1}{ \var{n2}}}} = \var{tval}\] to 3 decimal places.

(Using $s=\var{tpsd}$ in this formula. )

c)

Step 4: p value range.

As the degree of freedom is $\var{n1}+\var{n2}-2=\var{n-1}$ we use the $t_{\var{n-1}}$ tables. We have the following data from the tables:

p value	10%	5%	1%
Critical Value	1.717	2.074	2.819

We see that the $p$ value is less than 1%.

d)

Step 5: Conclusion

Hence there is strong evidence against $\operatorname{H}_0$ and so we reject $\operatorname{H}_0$.

There is evidence to suggest that average call times for cat lovers and dog lovers differ.

Question 3

The Newcastle Psychology department wants to estimate the mean IQ scores from students at Newcastle University.

A random sample of $\var{n}$ mathematics students gave a mean and standard deviation of $\var{m[s]}$ and $\var{sd[s]}$ respectively.

a)

Is the population variance known or unknown?

Unknown
Known

Expected answer:

Known
Unknown

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Score: 0/1

Unanswered

b)

Calculate a $\var{confl}$% confidence interval $(a,b)$ for the population mean:

$a=$ Round your answer to 2 decimal places. Expected answer:

$b=$ Round your answer to 2 decimal places. Expected answer:

Enter both to 2 decimal places.

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Score: 0/2

Unanswered

Advice

a)

The population variance is unknown. So we have to use the $t$ tables to find the confidence interval.

b)

We now calculate the $\var{confl}$% confidence interval:

As we have $\var{n}-1=\var{n-1}$ degrees of freedom, the interval is given by:

\[ \var{m[s]} \pm t_{\var{n-1}}\sqrt{\frac{\var{sd[s]}^2}{\var{n}}}\]

Looking up the t tables for $\var{confl}$% we see that $t_{\var{n-1}}=\var{invt}$ to 3 decimal places.

Hence:

Lower value of the confidence interval $=\;\displaystyle \var{m[s]} -\var{invt} \sqrt{\frac{\var{sd[s]} ^ 2} {\var{n}}} = \var{p}\var{lci}$ to 2 decimal places.

Upper value of the confidence interval $=\;\displaystyle \var{m[s]} +\var{invt} \sqrt{\frac{\var{sd[s]} ^ 2} {\var{n}}} = \var{p}\var{uci}$ to 2 decimal places.

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mathcentre: Parametric hypothesis testing for psychology

Question 1

a)

b)

c)

d)

Advice

a)

Step 1: Null Hypothesis

Step 2: Alternative Hypothesis

b)

c)

d)

Question 2

a)

b)

c)

d)

Advice

b)

Step 3 : Test statistic

c)

Step 4: p value range.

d)

Step 5: Conclusion

Question 3

a)

b)

Advice

a)

b)