Question

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. What is the p-value?

Answer 1

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = $5

For the alternative hypothesis,

µ < $5

number of samples taken = 10

Sample mean, x = (4 + 3 + 2 + 3 + 1 + 7 + 2 + 1 + 1 + 2)/10 = 2.6

To determine sample standard deviation, s

s = √(summation(x - mean)/n

n = 12

Summation(x - mean) = (4 - 2.6)^2 + (3 - 2.6)^2 + (2 - 2.6)^2 + (3 - 2.6)^2 + (1 - 2.6)^2 + (7 - 2.6)^2 + (2 - 2.6)^2 + (1 - 2.6)^2 + (1 - 2.6)^2 + (2 - 2.6)^2 = 30.4

s = √30.4/10 = 1.74

Since the number of samples is 10 and no population standard deviation is given, the distribution is a student's t.

Since n = 10,

Degrees of freedom, df = n - 1 = 10 - 1 = 9

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.6

µ = population mean = 5

s = samples standard deviation = 1.74

t = (2.6 - 5)/(1.74/√10) = - 4.36

We would determine the p value at alpha = 0.05. using the t test calculator. It becomes

p = 0.000912