OCR S3 2015 June — Question 4 9 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample confidence interval t-distribution
DifficultyStandard +0.3 This is a straightforward t-distribution confidence interval question with small sample size. Part (i) requires standard application of the t-interval formula with calculation of sample mean and standard deviation. Part (ii) involves working backwards from interval width to find the confidence level, requiring understanding of the relationship between critical value and interval width. While it involves multiple steps and careful calculation, it's a routine application of standard techniques with no novel insight required, making it slightly easier than average.
Spec5.05d Confidence intervals: using normal distribution

4 A set of bathroom scales is known to operate with an error which is normally distributed. One morning a man weighs himself 4 times. The 4 values for his mass, in kg , which can be considered to be a random sample are as follows. $$\begin{array} { l l l l } 62.6 & 62.8 & 62.1 & 62.5 \end{array}$$
  1. Find a \(95 \%\) confidence interval for his mass. Give the end-points of the interval correct to 3 decimal places.
  2. Based on these results, a \(y \%\) confidence interval has width 0.482 . Find \(y\).

Question 4:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\(\bar{x} = 62.5\)B1
\(\hat{\sigma} = 0.294(392)\)B1 or var=\(13/150 = 0.0866\ldots\) Allow 0.087 for this mark.
\(t = 3.182\)B1
\(\text{"62.5"} + \text{"3.182"} \times \text{"0.294"}/2\)M1 any t or z=1.96. \(\bar{x} \pm \frac{t\hat{\sigma}}{2}\)
\([62.032, 62.968]\)A1 Allow 4 or more dp.
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
"0.294" \(t/2\) oeB1ft Must be t. z=B0. If t written, but z used B0
\(= 0.241\) soiB1
1.637B1 Allow 1.63, 1.638, 1.64
80% CAOB1
## Question 4:

### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 62.5$ | B1 | |
| $\hat{\sigma} = 0.294(392)$ | B1 | or var=$13/150 = 0.0866\ldots$ Allow 0.087 for this mark. |
| $t = 3.182$ | B1 | |
| $\text{"62.5"} + \text{"3.182"} \times \text{"0.294"}/2$ | M1 | any t or z=1.96. $\bar{x} \pm \frac{t\hat{\sigma}}{2}$ |
| $[62.032, 62.968]$ | A1 | Allow 4 or more dp. |

### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| "0.294" $t/2$ oe | B1ft | Must be t. z=B0. If t written, but z used B0 |
| $= 0.241$ soi | B1 | |
| 1.637 | B1 | Allow 1.63, 1.638, 1.64 |
| 80% CAO | B1 | |

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4 A set of bathroom scales is known to operate with an error which is normally distributed. One morning a man weighs himself 4 times. The 4 values for his mass, in kg , which can be considered to be a random sample are as follows.

$$\begin{array} { l l l l } 
62.6 & 62.8 & 62.1 & 62.5
\end{array}$$

(i) Find a $95 \%$ confidence interval for his mass. Give the end-points of the interval correct to 3 decimal places.\\
(ii) Based on these results, a $y \%$ confidence interval has width 0.482 . Find $y$.

\hfill \mbox{\textit{OCR S3 2015 Q4 [9]}}