Test Quiz 5

A strength calculation on an old tube in a construction is to be performed. Because of corrosion and age diameters are quite ‘indeterminate’. Therefore several measurements are made of as well outer as inner diameter. The measurements of outer respectively inner diameter are independent of each other. The results are listed below: (all dimensions in mm)

Outer diameter, x: 44.9, 44.2 , 44.6, 44.8 , 44.0, 45.1

Inner diameter, y: 32.4, 32.5, 31.5, 32.2, 32.6, 31.7

From the data we get:

\[\left( {\bar x;{s_x}} \right) = (44.6;0.424)\,mm\quad \mbox{and} \quad \left( {\bar y;{s_y}} \right) = (32.15;0.451)\,mm\]

The outer diameter of the tube is as new 45 mm. The following test is performed:

\[\begin{array}{l} {H_0}:{\mu _x} = 45\\ \end{array}\]

At a 10% level of significance the result of this study is: (As well conclusion as argument must be correct):

The outer diameter is significantly different than the original, since the P-value is approximately 0.07

The t-statistic becomes: $t=\frac{44.6-45}{0.424/6}=-2.31$ and using a t-distribution with df=5 the P-value becomes $2\cdot P(t\leq -2.31)=0.0689$ found in R by \texttt{2*pt(-2.31,5)}.

If the data was put into R, the result could also be found as: > x <- c(44.9, 44.2, 44.6, 44.8, 44, 45.1) > t.test(x, mu = 45)

	One Sample t-test

data:  x
t = -2.3094, df = 5, p-value = 0.06896
alternative hypothesis: true mean is not equal to 45
95 percent confidence interval:
 	44.15476 45.04524
sample estimates:
mean of x 
 	44.6

Anyway, answer 1 is clearly the correct answer: The outer diameter is significantly smaller than the original, since the P-value is approximately 0.07

The outer diameter is not significantly smaller than the original, since the P-value is approximately 0.010

The outer diameter is significantly smaller than the original, since the P-value is approximately 0.035

The outer diameter is not significantly smaller than the original, since the P-value is approximately 0.035

The outer diameter is significantly smaller than the original, since the P-value is approximately 0.010

In a consumer survey performed by a newspaper, 20 different groceries (products) were purchased in a grocery store. Discrepancies between the price appearing on the sales slip and the shelf price were found in 6 of these purchased products.

A consumer claims that there generally is inconsistencies on 25% of the prices in the store. If the consumer is right, the probability that on the purchase of 20 different products there are inconsistencies on exactly 6 items is:

0.383

0.25

0.617

0.169

A consumer claims that there generally is inconsistencies on 25% of the pricess in the store. If the consumer is right, the probability that on the purchase of 20 different products there are inconsistencies on exactly 6 items is?

Let $X$ be the random number of inconsistensies out of the 20, so $X$ is binomial$(n,p)$-distributed with $n=20$ and $p=0.25$. We are asked to find:

\[P(X=6) = \left(\begin{array}{c}20\\6\end{array}\right)\cdot 0.25^6\cdot 0.75^{14}=0.1686\]

It is found in R, by: dbinom(6,20,0.25). Hence the answer is 4:

		0.169

0.831

At large university courses, it is the aim that the grades for passed students has the following distribution:


Grade	2	4	7	10	12
Percentage	10	25	30	25	10

The variance for the distribution of the grades for passed students is:

$9.5$

$0.95$

$7$

$5$

$2.85$

We have data on the difference between monthly average temperatures in the northern part of Sealand (2011-2014), and historical (1961-1990) correspondingly for the northern part of Sealand (for example, the month of January in 2012 was 2.1 degrees warmer than the average for January in the years 1961-1990). The average of the $n=48$ observations is $\bar{x}=1.363$ and the standard deviation is $s=1.521$. In the following it may be assumed that the observations are independent and comes from a normal distribution ($X_i\sim N(\mu,\sigma^2)$).

The following hypothesis should now be tested \begin{align} H_0:\quad \mu=0 \end{align} against the two-sided alternative.

On level $\alpha=0.05$, what is the conclusion of this hypothesis test? (conclusion as well as argument should be correct) ($df$, $df_1$ and $df_2$ denote degrees of freedom for sampling distributions where relevant)

$H_0$ cannot be rejected, since $t_{obs}= \frac{\bar{x}}{s^2/\sqrt{48}}=4.08>t_{1-\alpha}=1.68$ ($df=47$)

$H_0$ is rejected, since $F_{obs}= \frac{\bar{x}^2}{s^2/\sqrt{48}}=5.56>F_{1-\alpha}=3.26$ ($df_1=4$ and $df_2=12$)

$H_0$ is rejected, since $t_{obs}=\frac{\bar{x}}{s^2/\sqrt{47}}=4.04>t_{1-\alpha/2}=2.01$ ($df=47$)

$H_0$ cannot be rejected, since $F_{obs}=\frac{\bar{x}^2}{s^2/\sqrt{47}}=5.51<F_{1-\alpha/2}=8.75$ ($df_1=12$ and $df_2=4$)

$H_0$ is rejected, since $t_{obs}=\frac{\bar{x}}{s/\sqrt{48}}=6.21>t_{1-\alpha/2}=2.01$ ($df=47$)

This is a copy of the information from the previous question: We have data on the difference between monthly average temperatures in the northern part of Sealand (2011-2014), and historical (1961-1990) correspondingly for the northern part of Sealand (for example, the month of January in 2012 was 2.1 degrees warmer than the average for January in the years 1961-1990). The average of the $n=48$ observations is $\bar{x}=1.363$ and the standard deviation is $s=1.521$. In the following it may be assumed that the observations are independent and comes from a normal distribution ($X_i\sim N(\mu,\sigma^2)$).

What is the 95% confidence interval for the standard deviation $\sigma$?

$[1.58,3.54]$

$\bar{x}\pm \frac{s}{\sqrt{48}}\cdot t_{0.975}=[0.92,1.80]$ (df is 47)

$\left[\sqrt{\frac{47\cdot s^2}{\chi^2_{0.975}}}; \sqrt{\frac{47\cdot s^2}{\chi^2_{0.025}}}\right]= [1.27,1.91]$

(df is 47)

$\bar{x}\pm \frac{s}{\sqrt{47}}\cdot t_{0.95}= [0.99,1.74]$ (df is 47)

$s^2\pm \frac{\bar{x}}{47}\cdot \chi^2_{0.95}=[0.42,4.20]$ (df is 48)

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