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\( \begingroup \)

The management of {sc[s]} wants to {dothis[s]}.

A random sample of {spec} $\var{n}$ {t[s]} gave a mean and standard deviation of  {p}$\var{m[s]}${units} and {p}$\var{sd[s]}${units}  respectively.

 

 

Is the population variance known or unknown? 

 Expected answer:

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

$a=\;${p} Expected answer: {units}          $b=\;${p} Expected answer: {units}

Enter both to 2 decimal places.

 

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

1.

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

2.

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 $=\;\displaystyle \var{m[s]} -\var{invt} \sqrt{\frac{\var{sd[s]} ^ 2} {\var{n}}} = \var{p}\var{lci}${units} to 2 decimal places.

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

 
\( \endgroup \)
\( \begingroup \)

A company {sc[s]} {dothis[s]} $\var{sd[s]}$ {units}.

A random sample of $\var{n}$ {t[s]} gives a mean  of $\var{m[s]}$ {units}. 

 

a)

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

$a=\;$ Expected answer: {units}          $b=\;$ Expected answer: {units}

Enter both to 2 decimal places.

 

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

{howwell[s]}

 Expected answer:

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a)

We use the z tables to find the confidence interval as we know the population variance.

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

Note that $z_{\var{confl}}=\var{zval}$ and the confidence interval is given by:

\[ \var{m[s]} \pm z_{\var{confl}}\sqrt{\frac{\var{sd2}}{\var{n}}}\]

Hence:

Lower value of the confidence interval $=\;\displaystyle \var{m[s]} -\var{zval} \sqrt{\frac{\var{sd2}} {\var{n}}} = \var{lci}${units} to 2 decimal places.

Upper value of the confidence interval $=\;\displaystyle \var{m[s]} +\var{zval} \sqrt{\frac{\var{sd2}} {\var{n}}} = \var{uci}${units} to 2 decimal places.

b)

Since $\var{aim}$ {doornot} {lies} in the confidence interval the answer is {Correct}.

 
\( \endgroup \)
\( \begingroup \)

{this} 

{claim}

{test}

A sample of {n} {things}

{resultis} £{m} with a standard  deviation of £{stand}.

Perform an appropriate hypothesis test to see if the claim made by the online flight company is substantiated (use a two-tailed test).

a)

Step 1: Null Hypothesis

$\operatorname{H}_0\;: \; \mu=\;$ Expected answer:

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;: \; \mu \neq\;$ Expected answer:

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Step 3: Test statistic

Should we use the z or t test statistic? Expected answer: (enter z or t)

Now calculate the test statistic = ? Expected answer: (to 3 decimal places)

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

c)

Step 4: p-value

Use tables to find a range for your $p$-value. 

Choose the correct range here for $p$ :

Expected answer:

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d)

Step 5: Conclusion

 

Given the $p$ - value and the range you have found, what is the strength of evidence against the null hypothesis?

 Expected answer:

Your Decision:

 Expected answer:

 

Conclusion:

 Expected answer:

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

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:

{table([['Critical Value',crit[0],crit[1],crit[2]]],['p value','10%','5%','1%'])}

We see that the $p$ value {pm[pval]}.


d)

Hence there is {evi1[pval]} evidence against $\operatorname{H}_0$ and so we {dothis} $\operatorname{H}_0$.

{Correctc}
\( \endgroup \)
\( \begingroup \)

{this} {stuff}

{claim}$\var{thismuch}${units} and {var} {thisvar}.

{test}

To investigate a sample of $\var{n}$ {things} {resultis} $\var{m}${units}. 

Perform an appropriate hypothesis test to see if the claim made by the customers is substantiated.

a)

Step 1: Null Hypothesis

$\operatorname{H}_0\;: \; \mu=\;$ Expected answer:

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;: \; \mu \lt\;$ Expected answer:

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Step 3: Test statistic

Should we use the z or t test statistic? Expected answer: (enter z or t)

Now calculate the test statistic = ? Expected answer: (to 3 decimal places)

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

c)

Step 4: p-value

Use tables to find a range for your $p$-value. 

Choose the correct range here for $p$ :

Expected answer:

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d)

Step 5: Conclusion

 

Given the $p$ - value and the range you have found, what is the strength of evidence against the null hypothesis?

 Expected answer:

Your Decision:

 Expected answer:

 

Conclusion:

 Expected answer:

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a)

Step 1: Null Hypothesis

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

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;: \; \mu \lt\;\var{thismuch}$

b)

We should use the z statistic as the population variance is known.

The test statistic:

\[z =\frac{ |\var{m} -\var{thismuch}|} {\sqrt{\frac{\var{thisvar}}{\var{n}}}} = \var{zval}\]

to 3 decimal places.

c)

{table([['Critical Value',crit[0],crit[1],crit[2]]],['p value','10%','5%','1%'])}

We see that the $p$ value {pm[pval]}.


d)

Hence there is {evi1[pval]} evidence against $\operatorname{H}_0$ and so we {dothis} $\operatorname{H}_0$.

{Correctc}
\( \endgroup \)
\( \begingroup \)

{this}

A random  sample of $\var{n1}$  {things} and $\var{n2}$  {things1} gave the following results in {units}.

{table([['Male',{m},{sd}],['Female',{m1},{sd1}]],[' ','Mean','Standard deviation'])}

Perform an appropriate hypothesis test to see if there is any difference between {that} between {things} and {things1} (the null and alternative hypotheses have been set out for you).

a)

Step 1: Null hypothesis

If $\mu_M$ is the mean for time spent by {things} and  $\mu_F$ is the mean for time spent by {things1} then you are given that:

$\operatorname{H}_0\;:\;\mu_M=\mu_F$. 

Step 2: Alternative Hypothesis

$\operatorname{H}_1\;:\;\mu_M \neq \mu_F$. 

 

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Step 3: Test statistic

Should we use the z or t test statistic? 

Expected answer:

Now calculate the pooled standard deviation: Expected answer: (to 3 decimal places)

 

Now calculate the test statistic = ? Expected answer:  (to 3 decimal places)

 

(Note that in this calculation you should use a value for the pooled standard deviation which is accurate to at least 5 decimal places and not the value you found to 3 decimal places above).

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c)

Step 4:  p-value range

Use tables to find a range for your p -value. 

Choose the correct range here for p :

Expected answer:

 

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d)

 Given the $p$ - value and the range you have found, what is the strength of evidence against the null hypothesis?

 Expected answer:

Your Decision:

 Expected answer:

 

Conclusion:

 Expected answer:

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

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:

{table([['Critical Value',crit[0],crit[1],crit[2]]],['p value','10%','5%','1%'])}

We see that the $p$ value {pm[pval]}.

d)

Step 5: Conclusion

Hence there is {evi1[pval]} evidence against $\operatorname{H}_0$ and so we {dothis} $\operatorname{H}_0$.

{Correctc}.
\( \endgroup \)