The Principal of SACRO Primary is concerned that the average speed of cars travelling along Princess Road exceeds the recommended speed of 60km/h. A random sample of 10 cars had their speeds measured by radar and the results (in km/h) are as follows:

81, 53, 62, 79, 52, 68, 61, 85, 67
Assuming the speed of cars travelling along this road is normally distributed, does this data provide sufficient to support the Principal’s claim at the 5% level of significance?

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To determine whether the data provides sufficient evidence to support the Principal's claim at the 5% level of significance, we need to perform a hypothesis test.

Step 1: Formulate the null and alternative hypotheses.
- Null hypothesis (H0): The average speed of cars traveling along Princess Road is not greater than 60 km/h.
- Alternative hypothesis (H1): The average speed of cars traveling along Princess Road is greater than 60 km/h.

Step 2: Set the significance level (α).
In this case, the significance level is given as 5%, which means α = 0.05.

Step 3: Conduct the hypothesis test.
We will use a one-sample t-test since we only have one sample of data.

Step 4: Calculate the test statistic and p-value.
- Calculate the sample mean (x̄) and sample standard deviation (s) of the speeds.
- Calculate the t-value using the formula: t = (x̄ - μ) / (s / sqrt(n)), where μ is the hypothesized mean (60 km/h) and n is the sample size.
- Determine the p-value associated with the calculated t-value.

Step 5: Make a decision.
If the p-value is less than the significance level (α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Let's calculate the test statistic and p-value using the given data:

Sample mean (x̄) = (81 + 53 + 62 + 79 + 52 + 68 + 61 + 85 + 67) / 9 = 67.22
Sample standard deviation (s) = sqrt(((81-67.22)^2 + (53-67.22)^2 + (62-67.22)^2 + (79-67.22)^2 + (52-67.22)^2 + (68-67.22)^2 + (61-67.22)^2 + (85-67.22)^2 + (67-67.22)^2)/9) = 10.90
t-value = (67.22 - 60) / (10.90 / sqrt(9)) = 2.40

Degrees of freedom (df) = n - 1 = 9 - 1 = 8
Using a t-table or a statistical software, we find that the p-value associated with a t-value of 2.40 and df = 8 is approximately 0.021.

Step 6: Make a decision.
Since the p-value (0.021) is less than the significance level (0.05), we reject the null hypothesis. There is sufficient evidence to support the principal's claim that the average speed of cars traveling along Princess Road exceeds 60 km/h at the 5% level of significance.

Remember, you can double-check the calculations using statistical software or a t-table to ensure accuracy.