Standard +0.3 This is a standard two-sample t-test with all summary statistics provided. Students must state hypotheses (one-tailed), calculate the pooled standard error, compute the test statistic, and compare to critical value at 1% significance. While it requires careful execution of multiple steps, it's a routine application of a well-practiced procedure with no conceptual surprises—slightly easier than average due to straightforward setup and given values.
3. A grocer believes that the average weight of a grapefruit from farm \(A\) is greater than the average weight of a grapefruit from farm \(B\). The weights, in grams, of 80 grapefruit selected at random from farm \(A\) have a mean value of 532 g and a standard deviation, \(s _ { A }\), of 35 g . A random sample of 100 grapefruit from farm \(B\) have a mean weight of 520 g and a standard deviation, \(s _ { B }\), of 28 g .
Stating your hypotheses clearly and using a 1\% level of significance, test whether or not the grocer's belief is supported by the data.
Dependent on 1st M1. \(\pm\frac{(532-520)}{\text{"4.8117..."}}\); awrt 2.49
One tailed c.v. \(Z = 2.3263\) or CR: \(Z \geqslant 2.3263\); or p-value = awrt \(0.006 < 0.01\) or "0.994" \(> 0.99\)
B1
Critical value of 2.3263 or a correct probability comparison
Conclude: mean weight of grapefruit from farm \(A\) is greater than that of farm \(B\); or grocer's belief is correct
A1
Correct conclusion in context based on their \(z\)-value and their critical value, where \(
Alternative method: Let \(D = \bar{x}_A - \bar{x}_B\)
Answer
Marks
Guidance
\(2.3263 = \frac{D-0}{4.8117...}\)
dM1
Dependent on 1st M1; \(\frac{D}{\text{"4.8117..."}} = 2.3263/2.32/2.33\)
\(D = 11.193\); So \(D = \text{awrt}\ 11.2\)
A1, B1
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_A = \mu_B$, $H_1: \mu_A > \mu_B$ | B1 | If $\mu_1, \mu_2$ used must be clear which refers to which farm |
| $\text{s.e.} = \sqrt{\frac{35^2}{80} + \frac{28^2}{100}}\ \{= 4.81170448...\}$ | M1 A1 | Condone minor slips e.g. swapped $n$ or one s.d. and one variance. A1: s.e. = awrt 4.81 |
| $z = \frac{532-520}{\text{"4.8117..."}}\ ; = 2.4939...$ | dM1; A1 | Dependent on 1st M1. $\pm\frac{(532-520)}{\text{"4.8117..."}}$; awrt 2.49 |
| One tailed c.v. $Z = 2.3263$ or CR: $Z \geqslant 2.3263$; or p-value = awrt $0.006 < 0.01$ or "0.994" $> 0.99$ | B1 | Critical value of 2.3263 or a correct probability comparison |
| Conclude: mean weight of grapefruit from farm $A$ is greater than that of farm $B$; or grocer's belief is correct | A1 | Correct conclusion in context based on **their** $z$-value and **their** critical value, where $|c.v.| > 1$ |
**Alternative method:** Let $D = \bar{x}_A - \bar{x}_B$
| $2.3263 = \frac{D-0}{4.8117...}$ | dM1 | Dependent on 1st M1; $\frac{D}{\text{"4.8117..."}} = 2.3263/2.32/2.33$ |
|---|---|---|
| $D = 11.193$; So $D = \text{awrt}\ 11.2$ | A1, B1 | |
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3. A grocer believes that the average weight of a grapefruit from farm $A$ is greater than the average weight of a grapefruit from farm $B$. The weights, in grams, of 80 grapefruit selected at random from farm $A$ have a mean value of 532 g and a standard deviation, $s _ { A }$, of 35 g . A random sample of 100 grapefruit from farm $B$ have a mean weight of 520 g and a standard deviation, $s _ { B }$, of 28 g .
Stating your hypotheses clearly and using a 1\% level of significance, test whether or not the grocer's belief is supported by the data.\\
\hfill \mbox{\textit{Edexcel S3 2014 Q3 [7]}}