Question

5. A start-up company in Dublin offers online training. They advertise their courses via Google. In 2014 they advertised 30 courses and obtained an average of 44.2 students per course with a standard deviation of 4.6 students. In 2015 they did not use Google, but recruited an average 41.5 students and a standard deviation of 6.1 students per course across a different set of 32 courses (a) Calculate the standard errors of the mean for 2014 and 2015. Why are the standard errors smaller than the standard deviations? 3 marks (b) The company wish to calculate a confidence interval for the difference between 2015 and 2014 and so help determine whether the Google advertising is recruit- ing them more students. What is a confidence interval? Discuss the factors that make confidence intervals smaller or wider [5 marks] (c) Compute a 95% confidence interval for the difference in students between 2014 and 2015. You should assume unequal variances and use the formula 112 72 df n n1 to help you in your analysis. Discuss your answer in the context of the question [5 marks] (d) A statistician suggests that this analysis might be invalid because it breaks one of the key assumptions of the confidence interval. Which assumption might be broken here? How might they fix such a problem? 2 marks]
please show clearly all the calculations and indicate if you have used statisical tables (and which specific table used, eg. “percentage points of the x2 distribution), thank you!

Answer 1

(a) Standard error = $\dpi{100} s/\sqrt{n}$

For 2014 , n=30 , s= 4.6

Standard error = 0.8398

For 2015 , n=32 , s= 6.1

Standard error = 1.0783

Since standard error is obtained by diving standard deviation by square root of sample size , therefore standard error are smaller.

(b) $\dpi{100} (1-\alpha )100%$ % confidence interval for difference in mean number of students in 2014 and 2015 means we are $\dpi{100} (1-\alpha )100%$ % confident that the actual difference between mean number of students in two years lies in the interval , which is given by

$\dpi{100} (\bar{x1}-\bar{x2})\pm t_{c}*\sqrt{s1^{2}/n1+s2^{2}/n2}$

where tc is the critical value corresponding to $\dpi{100} (1-\alpha )100%$ % confidence , For higher confidence tc is higher , resulting in wider confidence interval.

When the sample sizes is larger , margin of error $\dpi{100} t_{c}*\sqrt{s1^{2}/n1+s2^{2}/n2}$ becomes smaller , resulting in narrow / small confidence interval.

(c) Calculation of df

$\dpi{100} (s1^{2}/n1+s2^{2}/n2)^{2}=3.49$

$\dpi{100} (s1^{2}/n1)^2/(n1-1)+(s2^{2}/n2)^{2}/(n2-1)=0.0608$

df = 57

tc = 2.002 (from t critical value table )

therefore 95% confidence interval is

$\dpi{100} (44.2-41.5)\pm 2.002*\sqrt{4.6^{2}/30+6.1^{2}/32}$

$\dpi{100} =2.7\pm 2.74$

=(-0.04 , 5.44)

(d) key assumptions are

(i) population are normally distributed

(ii) observations are independent between and within samples

Since it is not mentioned whether populations are normally distributed , this assumption might be violated , which can be fixed by taking large samples .