Test Quiz 10

Four medical drugs with the same active ingredient are compared in terms of side effects. 200 persons participate in a study where the side effects for the individual person is categorized as none / light / serious. The results of the investigation is shown in the table below:

Category	A	B	C	D	Total
No side effects	5	3	10	2	20
Light side effects	32	32	21	40	125
Serious side effects	13	15	19	8	55
Sum	50	50	50	50	200

If the entire data material is used, a 95% confidence interval for the proportion of serious side effects becomes:

$0.300 < {p_{serious}} < 0.458$

$0.213 < {p_{serious}} < 0.337$

$0.058 < {p_{serious}} < 0.262$

$0.190 < {p_{serious}} < 0.310$

$0.213 < {p_{serious}} < 0.537$

If you did the previous exercise, the following is a repetition:

Four medical drugs with the same active ingredient are compared in terms of side effects. 200 persons participate in a study where the side effects for the individual person is categorized as none / light / serious. The results of the investigation is shown in the table below:

Category	A	B	C	D	Total
No side effects	5	3	10	2	20
Light side effects	32	32	21	40	125
Serious side effects	13	15	19	8	55
Sum	50	50	50	50	200

Apart from the overall test, an investigation is wanted to find out whether the proportion of serious side effects is different for drug D than for the others. Hence, the following hypothesis test is performed:

\[\begin{array}{l} {H_0}:{p_D} = {p_{others}}\\ {H_1}:{p_D} \not= {p_{others}} \end{array}\]

The test statistic and P-value for this hypothesis test are:

$Z = -2.10$ and P-value = $0.036$

$Z = -2.10$ and P-value = $0.982$

$Z = -2.99$ and P-value = $0.0014$

$Z = -2.10$ and P-value = $0.0035$

$Z = -2.99$ and P-value = $0.0028$

A company sells weather modification rockets aimed a triggering precipitations in clouds. In order to assess the efficiency of this process, measurements are carried out in two regions with similar climatological features over 100 different non-consecutive days.

The amount of rain on region 1 (without cloud insemination) and region 2 (with cloud insemination) were summarized by the proportion of rainy days $p$, the average rainfall amount for rainy days $m$ and the sample standard deviation of rainfall amount for rainy days $s$: $p_1=0.51$, $p_2=0.68$, $\bar{x}_1=5.97$mm, $\bar{x}_2=8.25$mm, $s_1=\sqrt{23}$mm, $s_2=\sqrt{21}$mm.

Which of the following statements about the process is mostly correct?

It is efficient at increasing the number of rainy days and the p-value equals $0.007$

We want to test the whether $p_2$ is significantly greater than $p_1$. This can, with the information given, be done using the two sample proportions hypothesis test. We use a non-directional alternative hypothesis (two-sided) and afterwards check if the observed test statistic is to the side where $p_2>p_1$.

We have that $n_1=n_2=100$ and we use Chapter~7. From this information, we calculate:

$\hat{p} = \frac{51+68}{200} = 0.595$ $z_{obs} = \frac{.51-.68}{\sqrt{\hat{p}(1-\hat{p}) \left(1/100 + 1/100 \right)}} = -2.44877$

Let $Z \sim N(0,1)$, then the $p$-value is $2P(Z < -2.44877) = 0.014$. This is the $p$-value, i.e. the probability of getting a value for the test statistic equal to the one we found here, or more extreme, if the null hypothesis is true. We need to check that $z_{obs}$ is such that $p_2>p_1$, which it is since $z_{obs}<0$ hence $p_1<p_2$.

		Correct answer is 1.

It is efficient at increasing the number of rainy days and the p-value equals $0.035$

It is efficient at increasing the number of rainy days and the p-value equals $0.05$

It is not efficient at increasing the number of rainy days and the p-value equals $0.0013$

Is is not efficient at increasing the number of rainy days and the p-value equals $0.13$

If you did the previous exercise, the following is a repetition:

A company sells weather modification rockets aimed a triggering precipitations in clouds. In order to assess the efficiency of this process, measurements are carried out in two regions with similar climatological features over 100 different non-consecutive days.

The amount of rain on region 1 (without cloud insemination) and region 2 (with cloud insemination) were summarized by the proportion of rainy days $p$, the average rainfall amount for rainy days $m$ and the sample standard deviation of rainfall amount for rainy days $s$: $p_1=0.51$, $p_2=0.68$, $\bar{x}_1=5.97$mm, $\bar{x}_2=8.25$mm, $s_1=\sqrt{23}$mm, $s_2=\sqrt{21}$mm.

Which of the following statements is mostly correct?

The process is tested efficient at increasing the amounts of rain falling during rainy days and the p-value is greater than $0.05$

The process is tested not efficient at increasing the amounts of rain falling during rainy days and the p-value greater than $0.05$

The process is tested efficient at increasing the amounts of rain falling during rainy days and the p-value equals $0.5$

The process is tested not efficient at increasing the amounts of rain falling during rainy days and the p-value equals $0.05$

The process is tested efficient at increasing the amounts of rain falling during rainy days and the p-value is less than $0.05$

We want to test whether $\mu_2$ is significantly larger than $\mu_1$. This can, with the information given, be done using the two-sample t-test. We use a non-directional alternative hypothesis and afterwards check if the observed test statistic is to side where $\mu_2>\mu_1$. First we need the sample sizes, which are the days with rain: $n_1 = p_1 \cdot 100 = 51 ~~\mbox{and}~~ n_2 = p_2 \cdot 100 = 68$ We then calculate the test statistic $t_{obs}$ in Python:

## the variables
n1 = 51 
n2 = 68
xbar1 = 5.97
xbar2 = 8.25
s1 = np.sqrt(23)
s2 = np.sqrt(21)

## the observed test statistic
t_obs = (xbar1-xbar2) / np.sqrt(s1**2/n1+s2**2/n2)
print(t_obs)
-2.615677

## the degrees of freedom
nu = (s1**2/n1+s2**2/n2)**2 / ((s1**2/n1)**2/(n1-1) +(s2**2/n2)**2/(n2-1))
print(nu)
105.1336

In order to find the $p$-value, we need to find the probability that a random variable $T$ following the $t$-distribution with the calculated degrees of freedom, is more extreme than the value of $t_{obs}$ just calculated. Since it is the alternative hypothesis is non-directional (two-sided) then

\[2 P(T > |t_{obs}|) = 2 (1 - P(T \leq t_{obs}))\]

which we can calculate in Python:

2 * (1 - (stats.t.cdf(abs(t_obs),nu)))
0.01021466

This is the $p$-value, i.e. the probability of getting a value for the test statistic equal to the one we found here, or more extreme, if the null hypothesis is true. We need to check that $t_{obs}$ is such that $\mu_2>\mu_1$, which it is since $t_{obs}<0$ hence $\mu_1<\mu_2$.

Thus we have enough evidence against the null hypothesis (at the usual significance level of $\alpha=0.05$) to reject this and conclude that the process is effective.

		Correct answer is 5.

In relation to medicine development, a number of candidates are designed on molecular level. For 832 drugs, the following correlation coefficients between the three properties, MW (molecular weigh), SURF (molecular surface area) and VOL (molecular volume) are achieved:

cors	SURF	VOL	MW
SURF	1	0.995	0.949
VOL	0.995	1	0.952
MW	0.949	0.952	1

If we fit the following regression model to the data:

$\mbox{MW}_i=\alpha + \beta\cdot \mbox{SURF}_i + \varepsilon_i$, how much of the variation in $\mbox{MW}_i$ will be explained?

90.1%

99.0%

1.0%

9.4%

None of the above

If you did the previous exercise, the following is a repetition:

In relation to medicine development, a number of candidates are designed on molecular level. For 832 drugs, the following correlation coefficients between the three properties, MW (molecular weigh), SURF (molecular surface area) and VOL (molecular volume) are achieved:

cors	SURF	VOL	MW
SURF	1	0.995	0.949
VOL	0.995	1	0.952
MW	0.949	0.952	1

Assume that we have fitted the following regression model to the data:

$\mbox{VOL}_i=\alpha + \beta\cdot \mbox{SURF}_i + \varepsilon_i$ If molecule A has a surface area (SURF) equal to 200 areal units larger than molecule B (that is\ $ \mbox{SURF}_A=\mbox{SURF}_B + 200$), the expected volume of molecule A is given with:

$\mbox{VOL}_A= 200 + \mbox{VOL}_B$

$\mbox{VOL}_A= 199 + \mbox{VOL}_B$

$\mbox{VOL}_A= 1+ \mbox{VOL}_B$

$\mbox{VOL}_A= \mbox{VOL}_B-199$

None of the above

In a study of 712 medicine candidates, the mean molecular weight was found to be 437 with standard deviation of 117.

Within what limits are 95% of this type of medicine expected to lie within?

$437\pm 1.645\frac{117}{437}$

$437\pm 1.96\frac{117^2}{437}$

$437\pm 1.96\cdot 117^2$

$437\pm 1.96\cdot 117$

$437\pm 1.645\cdot 117$

02402 · Test Quiz 10

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