Hypothesis Testing & SQC Practice

Module V: Statistical Techniques – III
Practice Questions with Solutions

Question 1 – Confidence Limits (Large Sample)

A random sample of 100 students has mean IQ 110 with standard deviation 15. Find 95% confidence limits for the population mean IQ.

Solution

For 95% confidence, $Z_{\alpha/2}=1.96$ .

\text{Confidence limits} = \bar{x} \pm Z_{\alpha/2}\frac{\sigma}{\sqrt{n}} = 110 \pm 1.96\cdot\frac{15}{\sqrt{100}} = 110 \pm 1.96\cdot1.5 = 110 \pm 2.94

\text{Lower limit} = 107.06, \quad \text{Upper limit} = 112.94

Question 2 – t-Test (One Sample)

A sample of 10 observations has mean 6.2 and standard deviation 1.5. Test whether the sample comes from a population with mean 5.5 at 5% level of significance.
(Given $t_{0.05,9}=2.262$ )

Solution

$H_0:\mu=5.5$ vs $H_1:\mu\neq5.5$

t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{6.2-5.5}{1.5/\sqrt{10}}=\frac{0.7}{0.4743}\approx1.476

Critical value $=2.262$ .
Since $|t|=1.476<2.262$ , accept $H_0$ .
The sample may come from a population with mean 5.5.

Question 3 – t-Test (Two Samples)

Two samples give the following results:

Sample	Size	Mean	Variance
A	10	15	4
B	12	14	5

Test whether the means differ significantly at 5% level.
(Given $t_{0.05,20}=2.086$ )

Solution

$H_0:\mu_1=\mu_2$ vs

Pooled variance:

s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{9\cdot4+11\cdot5}{20}=\frac{36+55}{20}=4.55

s_p=\sqrt{4.55}\approx2.133

t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{15-14}{2.133\sqrt{\frac{1}{10}+\frac{1}{12}}} =\frac{1}{2.133\sqrt{0.1833}}\approx\frac{1}{0.913}\approx1.095

$|t|=1.095<2.086$ → accept $H_0$ . Means do not differ significantly.

Question 4 – F-Test

Two samples of sizes 8 and 10 have variances 4.5 and 9.0 respectively. Test whether the variances are significantly different at 5% level.
(Given $F_{0.05}(7,9)=3.29$ )

Solution

$H_0:\sigma_1^2=\sigma_2^2$ vs

F=\frac{s_2^2}{s_1^2}=\frac{9.0}{4.5}=2 \quad\text{(larger variance in numerator)}

Critical value $F_{0.05}(9,7)\approx3.68$ (or given 3.29 for reverse).
Since $F=2<$ critical value, accept $H_{}$ . Variances are not significantly different.

Question 5 – Chi-Square Test (Goodness of Fit)

A die is thrown 120 times and the following frequencies are observed:

Face	1	2	3	4	5	6
Frequency	15	22	18	25	16	24

Test whether the die is fair at 5% level.
(Given $\chi^2_{0.05,5}=11.07$ )

Solution

Expected frequency under $H_0$ (fair die) $=120/6=20$ for each face.

\chi^2=\sum\frac{(O_i-E_i)^2}{E_i} =\frac{(15-20)^2}{20}+\frac{(22-20)^2}{20}+\frac{(18-20)^2}{20} +\frac{(25-20)^2}{20}+\frac{(16-20)^2}{20}+\frac{(24-20)^2}{20}

=\frac{25+4+4+25+16+16}{20}=\frac{90}{20}=4.5

$4.5<11.07$ → accept $H_0$ . The die is fair.

Question 6 – One-Way ANOVA

Three groups of students were given different teaching methods. Their scores are:

Method A	Method B	Method C
8	7	9
9	6	10
7	8	8
10	7	11

Perform one-way ANOVA and test at 5% level whether the methods differ.
(Given $F_{0.05}(2,9)=4.26$ )

Solution

\begin{align*} n_1=n_2=n_3=4,\quad N=12\\ \sum A=34,\ \bar{A}=8.5;\quad \sum B=28,\ \bar{B}=7;\quad \sum C=38,\ \bar{C}=9.5\\ \text{Grand total}=100,\quad \bar{x}=8.333 \end{align*}

\text{SSB}=4[(8.5-8.333)^2+(7-8.333)^2+(9.5-8.333)^2]=4[0.0278+1.777+1.361]=12.66

\text{SST}=\sum\sum x_{ij}^2 - \frac{(\sum\sum x_{ij})^2}{N}= (8^2+\cdots+11^2)-\frac{100^2}{12}=854-833.33=20.67

\text{SSW}=\text{SST}-\text{SSB}=20.67-12.66=8.01

\text{MSB}=\frac{12.66}{2}=6.33,\quad \text{MSW}=\frac{8.01}{9}=0.89

F=\frac{6.33}{0.89}\approx7.11

$7.11>4.26$ → reject $H_0$ . Teaching methods differ significantly.

Question 7 – $\bar{X}$ and $R$ Charts

The following are the means and ranges of 5 subgroups of size 4:

Subgroup	1	2	3	4	5
$\bar{X}$	12.5	11.8	13.0	12.2	12.7
$R$	2.1	1.8	2.5	1.9	2.0

Given $A_2=0.73$ , $D_3=0$ , $D_{4} = 2.28$ . Construct the control limits for and charts.

Solution

\bar{\bar{X}}=\frac{12.5+11.8+13.0+12.2+12.7}{5}=12.44

\bar{R}=\frac{2.1+1.8+2.5+1.9+2.0}{5}=2.06

$\bar{X}$ -Chart

\begin{align*} \text{UCL}&=\bar{\bar{X}}+A_2\bar{R}=12.44+0.73\times2.06=12.44+1.50=13.94\\ \text{CL}&=12.44\\ \text{LCL}&=12.44-1.50=10.94 \end{align*}

$R$ -Chart

\begin{align*} \text{UCL}&=D_4\bar{R}=2.28\times2.06=4.70\\ \text{CL}&=2.06\\ \text{LCL}&=D_3\bar{R}=0 \end{align*}

Question 8 – $p$ -Chart

In a manufacturing process, the fraction defectives in 6 samples of size 200 are:
0.02, 0.03, 0.015, 0.025, 0.04, 0.02.
Construct the control limits for the $p$ -chart.

Solution

\bar{p}=\frac{0.02+0.03+0.015+0.025+0.04+0.02}{6}=0.025

\begin{align*} \text{UCL}&=\bar{p}+3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}=0.025+3\sqrt{\frac{0.025\times0.975}{200}}\\ &=0.025+3\sqrt{0.000121875}=0.025+3\times0.01104=0.025+0.0331=0.0581\\ \text{CL}&=0.025\\ \text{LCL}&=0.025-0.0331=-0.0081\to0 \quad\text{(since proportion cannot be negative)} \end{align*}

Practice Tips

Always state $H_0$ and $H_1$ clearly and choose the correct critical value.
For small samples use $t$ -distribution; for variances use $F$ ; for frequencies use .

Question 1 – Confidence Limits (Large Sample)

Question 2 – t-Test (One Sample)

Question 3 – t-Test (Two Samples)

Question 4 – F-Test

Question 5 – Chi-Square Test (Goodness of Fit)

Question 6 – One-Way ANOVA

Question 7 – Xˉ\bar{X}Xˉ and RRR Charts

Question 8 – ppp-Chart

Practice Tips

Question 7 – $\bar{X}$ and $R$ Charts

Question 8 – $p$ -Chart