A random sample of 8 observations of a normal random variable X has mean $ \bar{x} $, where $ \bar{x} = 6.246 $ and $ \sum (x - \bar{x})^2 = 0.784 $. Test, at the 5% significance level, whether the population mean of $ X $ is less than 6.44.

A-Level CIE Further Maths 21.1 Poisson & Geometric Hypothesis Testing: A random sample of 8 observation

A random sample of 8 observations of a normal random variable X has mean $ \bar{x} $, where $ \bar{x} = 6.246 $ and $ \sum (x - \bar{x})^2 = 0.784 $ - CIE - A-Level Further Maths - Question 6 - 2016 - Paper 1

Question 6

$A-random-sample-of-8-observations-of-a-normal-random-variable-X-has-mean-$-\bar{x}-$,-where-$-\bar{x}-=-6.246-$-and-$-\sum-(x---\bar{x})^2-=-0.784-$-CIE-A-Level Further Maths-Question 6-2016-Paper 1.png$

A random sample of 8 observations of a normal random variable X has mean $ \bar{x} $, where $ \bar{x} = 6.246 $ and $ \sum (x - \bar{x})^2 = 0.784 $. Test, at... show full transcript

Worked Solution & Example Answer:A random sample of 8 observations of a normal random variable X has mean $ \bar{x} $, where $ \bar{x} = 6.246 $ and $ \sum (x - \bar{x})^2 = 0.784 $ - CIE - A-Level Further Maths - Question 6 - 2016 - Paper 1

Step 1

Estimate population variance:

96%

114 rated

Answer

The sample variance, ( s^2 ), can be estimated using the formula:

$s^2 = \frac{\sum (x - \bar{x})^2}{n - 1} = \frac{0.784}{8 - 1} = \frac{0.784}{7} \approx 0.112.$

Thus, we will use this variance in further calculations.

Step 2

State hypotheses:

99%

104 rated

Answer

We will conduct a one-tailed test where:

Null Hypothesis (( H_0 )): ( \mu \geq 6.44 )
Alternative Hypothesis (( H_1 )): ( \mu < 6.44 )

Step 3

Calculate value of test statistic:

96%

101 rated

Answer

Using the formula for the t-statistic:

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

Substituting in the values:

( \bar{x} = 6.246 )
( \mu_0 = 6.44 )
( s = \sqrt{0.112} \approx 0.334 )
( n = 8 )

We get:

$t = \frac{6.246 - 6.44}{0.334 / \sqrt{8}} \approx \frac{-0.194}{0.118} \approx -1.645.$

Step 4

State our correct tabular value:

98%

120 rated

Answer

Using a t-table for a one-tailed test with ( n-1 = 7 ) degrees of freedom and a significance level of 0.05, the critical value is approximately -1.895.

Step 5

Conclusion:

97%

117 rated

Answer

Since our calculated t-value of -1.645 is greater than -1.895, we do not reject the null hypothesis.

Thus, at the 5% significance level, we do not have sufficient evidence to conclude that the population mean of ( X ) is less than 6.44.

A random sample of 8 observations of a normal random variable X has mean \( \bar{x} \), where \( \bar{x} = 6.246 \) and \( \sum (x - \bar{x})^2 = 0.784 \) - CIE - A-Level Further Maths - Question 6 - 2016 - Paper 1

Worked Solution & Example Answer:A random sample of 8 observations of a normal random variable X has mean \( \bar{x} \), where \( \bar{x} = 6.246 \) and \( \sum (x - \bar{x})^2 = 0.784 \) - CIE - A-Level Further Maths - Question 6 - 2016 - Paper 1

Estimate population variance:

State hypotheses:

Calculate value of test statistic:

State our correct tabular value:

Conclusion:

Join the A-Level students using SimpleStudy...

Other A-Level Further Maths topics to explore