Question

5. At a conference in 2006, an executive from a brokerage firm in the money market, told a group of analysts that 70% of investors were confident in achieving their investment objectives. USB investor conducted a study with information from 300 investors and found that 67% were confident in achieving their investment objectives. Should the executive's claim be rejected, at 95% confidence? Interpret the result

Answer 1

Solution:

p₀ = 70% = 0.70

Claim : p = 0.70 or not

The null and alternative hypothesis can be written as

H₀ : p = 0.70 vs H₁ : p 0.70

sign in H1 indicates the TWO tailed test.

n = 300

Let denotes the sample proportion

= 67% = 0.67

We test the claim using 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

  = 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [(0.67* (1 - 0.67)/300]

=  0.0532

Now the confidence interval is given by

( - E)   ( + E)

(0.67 - 0.0532)   (0.67 - 0.0532)

0.6168 0.7232

Interval is (0.6168 , 0.7232 )

Observe that , the hypothetical value 0.70 is in the interval (0.6168 , 0.7232)

So, we fail to reject the null hypothesis H₀

We can there is no evidence to reject the claim that the proportion of investors that were confident in achieving their investment objectives is 70%.

i.e. The proportion of investors that were confident in achieving their investment objectives is not different from 70%.