(a) $P\;(z < -1.6789)$.

(b) $P\;(z > -1.6789)$.

(c) $P\;(|z| < 0.78)$.

(d) $P\;(-0.98 < z < 1.279)$

We will use a function named

0.04658576727700960.953414232722990.5646091248285340.736008412726515

Next, let us look at how to do reverse lookups.
**Example 2:** Find the z-score corresponding to
the following areas in the
standard normal distribution:

(a) Area in the upper tail is 0.0845 .

(b) Area in lower tail is 0.404.

(c) Want z corresponding to the central 48% area.

We will use a function named`qnorm`,
which returns the z-value based on accumulating areas from
the left-end (i.e., in the lower tail). Thus, for each
situation we must figure out the right input to give
`qnorm` so that it
gives us what we want.

(a) Area in the upper tail is 0.0845 .

(b) Area in lower tail is 0.404.

(c) Want z corresponding to the central 48% area.

We will use a function named

1.37542410526545-0.2430069674099820.643345405392917

where $k$ is the level of confidence we want.

1-sample proportions test with continuity correction
data: 0.6 * 900 out of 900, null probability 0.5
X-squared = 35.601, df = 1, p-value = 2.421e-09
alternative hypothesis: true p is not equal to 0.5
90 percent confidence interval:
0.5723185 0.6270697
sample estimates:
p
0.6

The confidence interval = [ 0.5723185 0.6270697 ]

To see a more general description of `prop.test`
use R's builtin help utility by typing

`?prop.test`

and "run" it.

and "run" it.

1-sample proportions test with continuity correction
data: x out of n, null probability p0
X-squared = 8.89, df = 1, p-value = 0.001434
alternative hypothesis: true p is greater than 0.55
95 percent confidence interval:
0.5723185 1.0000000
sample estimates:
p
0.6
The P-value= 0.001433675

(a) $P\;(t < 1.967)$ with 5 df.

(b) $P\;(t > 1.967)$ with 5 df.

(c) The $t$-value where 97.5% area is to the left, with 5 df.

(This is the same as the $t_5^*$ value for a 95% confidence interval.)

(d) The $t$-value where 97.5% area is to the left, with 23 df.

0.9468343550699760.05316564493002382.570581835636312.06865761041905

Suppose we treat those data as a random sample of cities drawn from a population that consists of all cities in the U.S. Compute a 90% confidence interval for the true mean latitude of U.S. cities.

One Sample t-test
data: x
t = 74.436, df = 58, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
90 percent confidence interval:
38.35037 40.11235
sample estimates:
mean of x
39.23136
The confidence interval = [ 38.35037 40.11235 ]

The P-value= 0.01147574

To see a more general description of `t.test`
use R's builtin help utility by typing

`?t.test`

and "run" it.

and "run" it.