Hypothesis Testing for Bus Routes and Birth Weights

This assignment covers two hypothesis testing scenarios. The first examines whether a new bus route should be added based on commuter survey results, using a 1% significance level. The second investigates if babies born to mothers who smoked have lower birth weights compared to other babies, employing a 5% significance level and analyzing data from a sample of male newborns.

Financial Statistics
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Student’s Number
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6th October 2017

A. The policy of a new Suburban Transit Authority in an Australian city says that it should
add a new bus route to the system if more than 45% of potential commuters indicate
that they will use that new route. A sample of 200 commuters revealed that 98 said
they would use the new bus route. Is there sufficient evidence, at the 1% significance
level, to conclude that the Transit Authority should add the new bus route?
It should be written in the below format
State the data type and parameter of interest:
The data is a categorical data type and the parameter of interest is the proportion of those
who indicate that they will use new route.
State the Null and Alternative hypotheses:
Ho: p=0.45
HA: p>0.45
State the test statistic and the distribution of its standardised value:
Test statistics:
State the level of significance:
This is tested at α = 0.01
P=0.01 significance level because we were provided with this value and it is one-tailed.
Find the critical value(s)(table value) and develop the decision rule:
Z-critical value is 2.33 (for a right one-tailed test); the null hypothesis is rejected when
the computed z value is greater than the critical value or when the p-value is less than α =
0.01 otherwise the null is accepted.
Perform the test using appropriate software. Obtain the value of the test statistic or the
p-value from the output and compare with the decision rule.
Pr(Z < z) = 0.8722 Pr(|Z| > |z|) = 0.2555 Pr(Z > z) = 0.1278
Ha: p < 0.45 Ha: p != 0.45 Ha: p > 0.45
Ho: p = 0.45
p = proportion(x) z = 1.1371
x .49 .0353483 .3989489 .5810511
Variable Mean Std. Err. [99% Conf. Interval]
One-sample test of proportion x: Number of obs = 200
. prtesti 200 0.49 0.45, level(99)
Sate your conclusion and the answer:

From the above table, we observe that the p-value is 0.1278; this value is greater than α =
0.01 hence we fail to reject the null hypothesis and conclude that the sample proportion of
those who said they would use the new bus route is not significantly greater than 45%. Thus
the company should not consider adding a new bus route to the system
B. The birth weights of male babies at a major hospital is known to follow a normal
distribution with a mean of 3050 grams and a standard deviation of 475 grams. A
random sample of 50 male babies, born to mothers who smoked during their
pregnancy, had a mean weight of 2835 grams. Based on this sample, and using a 5%
significance level, can we conclude that babies born to mothers who smoked during
pregnancy have lower birth weights than other babies?
State the data type and parameter of interest:
The data is a continuous (numerical) data type and the parameter of interest is the mean
weight of male babies
State the Null and Alternative hypotheses:
Ho:
μ=3050050 the company should not consider null hypothesiss 0.1278 ;this value is greater than he prod
HA: μ<3050
State the test statistic and the distribution of its standardised value.
State the level of significance:
α = 0.05 significance level because we were provided with this value and it is one-tailed.
Find the critical value(s)(table value) and develop the decision rule:
The z-critical value is 1.645 (for a left one-tiled test); the null hypothesis is rejected when the
computed z value is greater than the critical value or when the p-value is less than the α
=0.05
Perform the test using appropriate software. Obtain the value of the test statistic or the
p-value from the output and compare with the decision rule.

Pr(T < t) = 0.0012 Pr(|T| > |t|) = 0.0024 Pr(T > t) = 0.9988
Ha: mean < 3050 Ha: mean != 3050 Ha: mean > 3050
Ho: mean = 3050 degrees of freedom = 49
mean = mean(x) t = -3.2006
x 50 2835 67.17514 475 2700.006 2969.994
Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
One-sample t test
. ttesti 50 2835 475 3050
State your conclusion and the answer:
The p-value for the one-tailed (left direction) is 0.0012; this value is less than α = 0.05 thus
we reject the null hypothesis and conclude that there is evidence that babies born to mothers
who smoked during pregnancy have lower birth weights than other babies.