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میتونی لایو بذاری!

A clinic monitors the amount, $X$ milligrams per litre, of a certain chemical in the blood stream of patients. For patients who are taking drug $A$, it has been found that the mean value of $X$ is $0.336$. $A$ random sample of $100$ patients taking a new drug, $B$, was selected and the values of $X$ were found. 

The results are summarised below.

$n = 100$, $\sum x  = 43.5$, $\sum {{x^2}}  = 31.56$.

a) Test at the 1% significance level whether the mean amount of the chemical in the blood stream of patients taking drug $B$ is different from that of patients taking drug $A$.

b) For the test to be valid, is it necessary to assume a normal distribution for the amount of chemical in the blood stream of patients taking drug $B$? Justify your answer.

پاسخ تشریحی :
نمایش پاسخ

a) $\overline x  = 43.5/100 = 0.435$

$s = \sqrt {\frac{{100}}{{99}}}  \times \sqrt {\frac{{31.56}}{{100}} - {{0.435}^2}} \,\left( { = 0.3573} \right)$

or $Var{\text{ }}\left( { = 0.128} \right)$ or $1/99\left( {31.56 - {{\left( {43.5} \right)}^2}/100} \right)$

${H_0}:$ Pop mean (for $B$) $ = 0.336$

${H_1}:$ Pop mean (for $B$) $ \ne 0.336$

$\frac{{0.435 - 0.336}}{{\frac{{''0.3573''}}{{\sqrt {100} }}}}$

$ = 2.77{\text{ }}\left( {3{\text{ }}sfs} \right)$

${Z_{crit}} = 2.576$

(or $2.326$ consistent with 1-tail test)

Valid comparison with z-value

Evidence that $B$ amounts diff from $A$

b) Must state or imply “No” to score these marks

$n$ large

$\overline X $ approx normally distr or CLT applies

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