Question

Question 1

For smartphone, the battery performance is one of the important technical specifications. A company has designed a version of smartphone and has done some tests on the battery life. Let X denotes the amount of time that the fully charged battery can last for Internet surfing. The unit of X is hour and it is omitted in the following discussion.

(a) Suppose we know that X is normally distributed with mean μ=10 and standard deviation σ=1.2, i.e., X~N(10,〖1.2〗^2). For a randomly picked phone, what is the probability that its battery can last for longer than 9.3 hours for internet surfing?

(b) Suppose we know that X~N(10,〖1.2〗^2). If we randomly pick 36 phones and calculate the average of battery time for internet surfing, which is denoted as X ̅. Then, what is the value of P(X ̅>9.3)?

(c) Suppose we don’t know the distribution of X, but only know that its mean μ=10 and its standard deviation σ=1.2. Will the result in (b) be affected seriously? Why?

(d) Suppose we know nothing about X and take a random sample to infer the mean of X. If the sample size is n=36, and after calculation we find the sample mean is X ̅=9.8 and the sample standard deviation is s=1.3. Construct a 95% confidence interval estimate of the mean of X.

(e) Suppose we know that the standard deviation of X is σ=1.2. Based on the same sample in (d), what is the 95% confidence interval estimate of the mean of X? What is the sampling error in this case? If we want to reduce the sampling error by half, how many additional phones are required?

Answer 1

(a)

Probability that its battery can last for longer than 9.3 hours for internet surfing = P(X > 9.3)

= P[Z > (9.3 - 10)/1.2]

= P[Z > -0.58]

= 0.7190

(b)

By Central limit theorem,

will follow Normal distribution with mean = 10, and standard deviation = 1.2/ = 0.2

P( > 9.3)

= P[Z > (9.3 - 10)/0.2]

= P[Z > -3.5]

= 0.9998

(c)

No, the result in (b) would remain same. By Central limit theorem, the distribution of will always follow Normal distribution irrespective of the distributuon of X.

(d)

Since, we do not know the population standard deviation, we will use t statistic to construct the 95% confidence interval estimate.

Degree of freedom = n-1 = 36-1 = 35

Critical value of t at df = 35 and 95% confidence interval is 2.03

Standard error of mean = s / = 1.3/ = 0.2167

95% confidence interval estimate of the mean of X is,

(9.8 - 2.03 * 0.2167 , 9.8 + 2.03 * 0.2167)

(9.36 ,  10.24)

(e)

Since, we know the population standard deviation, we will use z statistic to construct the 95% confidence interval estimate.

Critical value of z at 95% confidence interval is 1.96

Standard error of mean = σ / = 1.2/ = 0.2

The sampling error is 0.2 hours

95% confidence interval estimate of the mean of X is,

(9.8 - 1.96 * 0.2 , 9.8 + 1.96 * 0.2)

(9.408 ,  10.192)

To reduce the sampling error by half, we need to increase the denominator from to 2 which is

Thus, Sample size to be increased by 4 * 36 = 144

Number of additional phones required = 144 - 36 = 108