4. A small electronics store has begun to advertise in the local newspaper. Befo
ID: 3339583 • Letter: 4
Question
4. A small electronics store has begun to advertise in the local newspaper. Before advertising, the long term mean weekly sales were $9820. A random sample of 10 weeks while the newspaper ads were running had a sample mean of $10,960 and a sample standard deviation of $1580. From previous data, the distribution of weekly sales is known to be approximately a normal distribution. At the 5% significance, does the data provide sufficient evidence to claim that the population mean weekly sales is now more than $9820. (a) State the null and alternate hypotheses. (b) What is the level of significance? (c) Is this a right-tailed, left-tailed, or two-tailed test? (d) List each assumption and show information to determine if the assumptions are met for testing this hypothesis. (e) Which of the following test procedures is used to test the hypothesis stated in part (a)? NOTE: COMPUTATION OF THE P-VALUE NOT REQUIRED a. One mean Z test b. One Mean-t test c. Cannot do test (State why) (1 If the p-value is 0.0242, is Ho rejected or not? (9) Write the interpretation of the results of the hypothesis test.Explanation / Answer
we use Z tets Statistic
Z = X- U / S.D / Sqrt(n)
a)
Ho:U>9820
H1:U<=9820
b) level of significance alpha =0.05
c) right tailed test
d)
n=50; x=10960; S.D= 1580
n> 30 So, we use Z tets Statistic
Z = X- U / S.D / Sqrt(n)
e)
Z = 10960 - 9820 / 1580 / Sqrt(50) = 5.1019
| Z cal | = 5.1019
| Z tab | at 5% LOS is = 1.96
We got | Z tab | < | Z cal|
f)
reject Ho
g)reject h0 as there is a sufficent evidence to say that pouulation mean is greater than $9800
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