interactive ppt (confidence interval)

How to Navigate

In this session, you will learn about the construction of a Confidence Interval.You can also click on the words underlined in blue, buttons below to go to the pages regarding these distributions, or on the navigation buttons.There is also a quiz at the end to test your understanding!

QUIZ

Confidence Interval

Confidence Interval Z Values Sample Sizet Values

Confidence IntervalThe confidence interval refers to a range of values constructed from sample data so that the population parameter is likely to occur within that range at a specified probability (level of confidence, 1-α).Confidence interval is represented by:

or

Where = or = ; = or =

QUIZ

± ±

Note: Z and t values in Confidence Interval should be read from the table. No

formula is required.

Confidence Interval


If σ is not known, you may use the sample standard

deviation, s.

Z Values Before reading the tables to get Z/t values, first decide the distribution it should follow (refer L7).

QUIZ

Example 1

Z value e.g. 95% Confidence level

Step 1. 95% ÷ 2 = 47.5% = 0.475

Step 2. Find probability closest to 0.475 inside normal table.

Step 3. Read Z value. Z = 1.96

Confidence Interval


Z Values Before reading the tables to get Z/t values, first decide the distribution it should follow (refer L7).

QUIZ

Example 2

Z value e.g. 90% Confidence level

Step 1. 90% ÷ 2 = 45% = 0.45

Step 2. Find probability closest to 0.45 inside normal table. You’ll notice that 0.4505 and 0.4495 is equally close to 0.45.

Step 3. Read both Z values. Z = 1.64 or Z = 1.65. You can also take the average of these two numbers to obtain 1.645.

Confidence Interval


0.4505 and 0.4495 are equally close because:

0.4505 – 0.45 = 0.00050.45 - 0.4495 = 0.0005

Click on Example 2 if you are unsure

why they are ‘equally close’.

t Values Before reading the tables to get Z/t values, first decide the distribution it should follow (refer L7).

QUIZ

t value e.g. 95% Confidence level

n = 8

Step 1. Find degrees of freedom = n – 1 = 8 – 1 = 7

Step 2. Read the column of ‘Confidence Level’

Step 3. Read the row that corresponds to the degrees of freedom. Therefore, t = 2.365

Confidence Interval


t Values Before reading the tables to get Z/t values, first decide the distribution it should follow (refer L7).

QUIZ

t value e.g. 99% Confidence level

n = 11

Step 1. Find degrees of freedom = n – 1 = 11 – 1 = 10

Step 2. Read the column of ‘Confidence Level’

Step 3. Read the row that corresponds to the degrees of freedom. Therefore, t = 3.169

Confidence Interval


Sample Size Determination• ε is the maximum tolerant sampling error. • Z is the z score that corresponds to a particular level of

confidence. • σ refers to the population standard deviation. If σ is not

known, you may use the sample standard deviation, s.

n =

Confidence Interval


Note: ALWAYS use z distribution. We will NEVER use t distribution in sample size determination.

ALWAYS round UP your answers. E.g. if n = 123.1 we will round UP the answer to 124.

QUIZ

QUIZ 1

A stationery store wishes to estimate the average retail value of bookmarks that it has in its inventory. A random sample of 20 bookmarks indicates an average value of $1.67 and a standard deviation of $0.32. Set up a 95% confidence interval estimate of the population average value of all bookmarks that are in its inventory.

Confidence Interval


Click on the screen for

guided steps

Step 1: Write down the given variables. Determine the distribution.

Step 2: Read the table to get the Z/t values.

Step 3: Substitute the values into the CI formula.

n = 20, = 1.67, s = 0.32, CI = 95%Since is unknown and n < 30, we use the t

distribution.

Degrees of freedom = n – 1 = 20 – 1 = 19t = 2.093

Confidence Interval = = 1.67 ± 2.093

= (1.5202, 1.8198)

QUIZ 2A stationery store wishes to estimate the average retail value of bookmarks that it has in its inventory out of 300 bookmarks. A random sample of 20 bookmarks indicates an average value of $1.67 and a standard deviation of $0.32. Set up a 95% confidence interval estimate of the population average value of all bookmarks that are in its inventory.

Confidence Interval



guided steps




N = 300, n = 20, = 1.67, s = 0.32, CI = 95%Since is unknown and n < 30, we use the t

distribution.

Degrees of freedom = n – 1 = 20 – 1 = 19t = 2.093

Since N is known, check for finite multiplier. = = 0.067 > 0.05 Therefore FM required


= (1.5251, 1.8149)

QUIZ 3A stationery store wishes to estimate the average retail value of bookmarks that it has in its inventory. A random sample of 20 bookmarks indicates an average value of $1.67 and a standard deviation of $0.32. The management of the store wishes to estimate the average retail value of bookmarks that it has in its inventory to within $0.08 with 90% confidence. What sample size is needed?

Confidence Interval



guided steps

90% CI Always use Z hence, Z = 1.645

= 0.08, s = 0.32Since σ is not known, use s instead.

n = = = 43.2964 ≈ 44 (Round UP)

QUIZ 4A stationery store wishes to estimate the average retail value of bookmarks that it has in its inventory out of 500 bookmarks with a mean of $1.70 and a standard deviation of $0.30. A random sample of 20 bookmarks indicates an average value of $1.67 and a standard deviation of $0.32. Set up a 95% confidence interval estimate of the population average value of all bookmarks that are in its inventory. Confidence Interval



guided steps




N = 500, µ = 1.70, = 0.30, n = 20, = 1.67, s = 0.32, CI = 95%

Since is known, we use the Z distribution.

95% ÷ 2 = 47.5%Z = 1.96

Since N is known, check for finite multiplier. = = 0.04 < 0.05 Therefore FM not required


= (1.5385, 1.8015)

QUIZ 5

A stationery store wishes to estimate the average retail value of bookmarks that it has in its inventory. A random sample of 50 bookmarks indicates an average value of $1.67 and a standard deviation of $0.32. Set up a 95% confidence interval estimate of the population average value of all bookmarks that are in its inventory.

Confidence Interval



guided steps




n = 50, = 1.67, s = 0.32, CI = 95%Since is unknown but n > 30, we use the Z

distribution.

95% ÷ 2 = 47.5%Z = 1.96


= (1.5813, 1.7587)

Summary• Always decide which distribution the sampling mean follows. • The Z/t values are read from the distribution tables. • Always remember to check if the finite multiplier is required if

N is known.• If σ is not known, use the sample standard deviation, s• When finding sample size, always use Z and always round up

my answer to a whole number.

Confidence Interval


QUIZ

± ± n =

interactive ppt (confidence interval)

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